For part a) Perform an inversion centred at $A$ with radius $\sqrt{AH*AD}$. You get the following equivalent problem. Inverted problem wrote: Let $\Delta ABC$ be a triangle and let a line parallel to $BC$ through $H$ intersect $\odot(AHB)$ and $\odot(AHC)$ at $G,I$. Let the tangent to $\odot(AHB)$ at $G$ meet $AB$ at $J$ and let the tangent to $\odot(AHC)$ at $I$ meet $AC$ at $K$. Prove that $JK$ is tangent to $\odot(ABC)$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.342914317675936, xmax = 42.65305614028167, ymin = -22.616358814798893, ymax = 11.060850247371606; /* image dimensions */ /* draw figures */ draw((7.755946918803331,4.973217220817929)--(4.584179684784288,-5.207764024181408), linewidth(2.8)); draw((4.584179684784288,-5.207764024181408)--(17.46,-4.72), linewidth(2.8)); draw((17.46,-4.72)--(7.755946918803331,4.973217220817929), linewidth(2.8)); draw((5.861813042443863,-1.106718678607489)--(17.46,-4.72), linewidth(1.2)); draw((4.584179684784288,-5.207764024181408)--(11.258778909391829,1.4742966060290714), linewidth(1.2)); draw((7.755946918803331,4.973217220817929)--(8.136526761929048,-5.073193405516467), linewidth(1.2)); draw(circle((1.4458832087446238,1.3544903965374384), 7.274069552442223), linewidth(1.2) + linetype("2 2")); draw(circle((14.321703523960336,1.84225442071885), 7.27406955244223), linewidth(1.2) + linetype("2 2")); draw((1.8524676543225498,-13.976222393564713)--(18.673282953753546,-5.931928162659889), linewidth(1.2)); draw(circle((10.894243394842997,-1.5890371999009185), 7.274069552442226), linewidth(1.2) + linetype("2 2")); draw((7.755946918803331,4.973217220817929)--(-4.864180501314085,-2.2642364277430493), linewidth(1.2)); draw((7.755946918803331,4.973217220817929)--(14.032539870882664,-8.151291620619773), linewidth(1.2)); draw((5.861813042443863,-1.106718678607489)--(8.136526761929048,-5.073193405516467), linewidth(1.2)); draw((8.136526761929048,-5.073193405516467)--(11.258778909391829,1.4742966060290714), linewidth(1.2)); draw((-4.864180501314085,-2.2642364277430493)--(4.584179684784288,-5.207764024181408), linewidth(1.2)); draw((4.584179684784288,-5.207764024181408)--(14.032539870882664,-8.151291620619773), linewidth(1.2)); draw((14.032539870882664,-8.151291620619773)--(20.887460129117336,-1.2887083793802327), linewidth(1.2)); draw((17.46,-4.72)--(18.673282953753546,-5.931928162659889), linewidth(1.2)); draw((4.584179684784288,-5.207764024181408)--(1.8524676543225498,-13.976222393564713), linewidth(1.2)); draw((-4.864180501314085,-2.2642364277430493)--(1.8524676543225498,-13.976222393564713), linewidth(1.2)); draw((20.887460129117336,-1.2887083793802327)--(18.673282953753546,-5.931928162659889), linewidth(1.2)); draw((-4.864180501314085,-2.2642364277430493)--(20.887460129117336,-1.2887083793802327), linewidth(1.2)); /* dots and labels */ dot((7.755946918803331,4.973217220817929),dotstyle); label("$A$", (7.883076282583302,5.298967870953055), NE * labelscalefactor); dot((4.584179684784288,-5.207764024181408),dotstyle); label("$B$", (4.571649629469172,-6.224796881884048), NE * labelscalefactor); dot((17.46,-4.72),dotstyle); label("$C$", (17.585556376207705,-4.403512222671288), NE * labelscalefactor); dot((8.011639813901622,-1.7764724035616408),linewidth(4pt) + dotstyle); label("$H$", (8.644704412799552,-1.555685300993153), NE * labelscalefactor); dot((8.136526761929048,-5.073193405516467),linewidth(4pt) + dotstyle); label("$D$", (8.280447480956997,-4.800883421044981), NE * labelscalefactor); dot((11.258778909391829,1.4742966060290714),linewidth(4pt) + dotstyle); label("$E$", (11.525645601008845,1.490827219871828), NE * labelscalefactor); dot((5.861813042443863,-1.106718678607489),linewidth(4pt) + dotstyle); label("$F$", (6.194248689495096,-0.9596285034326133), NE * labelscalefactor); dot((-4.864180501314085,-2.2642364277430493),linewidth(4pt) + dotstyle); label("$G$", (-5.495087395997784,-2.416656230802822), NE * labelscalefactor); dot((20.887460129117336,-1.2887083793802327),linewidth(4pt) + dotstyle); label("$I$", (21.26123996116439,-1.4563425013997298), NE * labelscalefactor); dot((1.8524676543225498,-13.976222393564713),linewidth(4pt) + dotstyle); label("$J$", (2.021851106571291,-14.437134981607041), NE * labelscalefactor); dot((18.673282953753546,-5.931928162659889),linewidth(4pt) + dotstyle); label("$K$", (18.843898504391074,-6.589053813726601), NE * labelscalefactor); dot((14.032539870882664,-8.151291620619773),linewidth(4pt) + dotstyle); label("$A'$", (14.174786923500148,-7.880510208441103), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Let $A'$ be the $A$-antipode w.r.t $\odot(ABC)$. Notice that $G-B-A'$ collinear . Similiarly $I-C-A'$ collinear. Now note that $AI$ is a diameter of $\odot(AHC)$. Well known fact that radius of $\odot(ABC)$ is equal to radius of $\odot(AHC) \implies \overline{AI}=\overline{AA'}$ . Similiarly $\overline{AG}=\overline{AA'}$. Now note that $\angle JA'G=\angle JGA'=90^\circ -\angle AGB =90^\circ-\angle AA'B \implies JA'$ is tangent to $\odot(ABC)$ . Similiarly $KA'$ is tangent to $\odot(ABC)$. But note that since $\angle JA'A+\angle KA'A=90^\circ+90^\circ=180^\circ \implies J-K-A'$ collinear which implies the result. $\blacksquare$.

For a) draw $AT \perp BC$ first For b) Let $EF \cap MN =X,$ then we need to prove $(X,K;M,N)=-1$ It reduces to prove the following problem : $\triangle ABC,$ orthic triangle $DEF.$ $M, N$ is the reflection of $D$ with $AB, AC.$ $MN \cap BC =X$ then $(X,D;B,C)=-1$ It suffices to prove $M$ and $N$ lies on $EF$ which is simple

General problem. Let $DEF$ be pedal triangle of any point $J$ with respect to triangle $ABC$. $A^*$ is isogonal conjugate of $A$ with respect to triangle $DEF$. $M$, $N$ are projections of $A^*$ on the lines $DE$, $DF$ respectively. $EF$ cuts $MN$ at $P$. $PA^*$ cuts $DA$ at $X$. $K$ is projection of $A$ on the line $EF$. $DK$ cuts $MN$ at $Q$. Prove that reflection of $X$ in line $MN$ lies on circle $(DPQ)$.

Solution for part a). Let $T$ be the projection of $A$ on $EF$. Clearly, $ATFN$ is cyclic. Since $AB$ is the external bisector of $\angle DFE$ we easily get that $ATFN$ is a cyclic kite (in fact, $AMDN$ is a cyclic kite as well), hence $N,\ P$ and $T$ are collinear. Similarly, $ATEM$ is a cyclic kite and $M,\ Q$ and $T$ lie on the same line. We deduce that $$\angle PAT=\angle NAF=\angle PTF$$so $EF$ touches $(PAQ)$ at $T$. Solution for part b). Let $S$ be the second intersection point of $(APQ)$ and $(MDN)$. Because $\angle TSA=90^\circ=\angle DSA$ we infer that $D,\ T$ and $S$ are collinear. Let $R=\overline{AS}\cap \overline{MN}$. Together with $RK\perp AD$ we conclude that $K$ is the orthocenter of $\bigtriangleup RAD$, therefore $$\angle DKR=180^\circ-\angle RAD=180^\circ-\angle RLA=\angle RLD$$thus $DLKR$ is cyclic. It suffices to show that $EF$ passes through $R$, but we know that $$\angle ASN=\angle AMN=\angle ANM$$so $AN$ is tangent to $(NSR)$, thus $$AT^2=AN^2=AS\cdot AR$$from which $\angle ATR=90^\circ$ and the desired collinearity follows. $\blacksquare$

Yahhh! A solution by Barycentric Coordinates

Solution for part a) Obviously, $\widehat {ABN}=\widehat {AHN}=90^{\circ}, \widehat {ANB}=180^{\circ}-\widehat {AHB}=\widehat {ATB},\widehat {ABT}=90^{\circ} $. This $\implies $ $N $ is the reflection of $T $ by $B\implies \widehat {PTB}=\widehat {PNB}=\widehat {PAN}=\widehat {TAB} $. Similarly, $\widehat {QTC}=\widehat {TAC}\implies\widehat {PTQ}=180^{\circ} $ Solution for part b) Applying homothety of center $T $ and ratio $\tfrac 12$ it is sufficient to prove this lemma. Lemma- Let $\triangle IBC $ be inscribed in $(L) $ has $(T) $ at its incircle. Let $H $ be the midpoint of $BC$ and $O\neq I $ be the intersection of $IT $ and $L $. Assume that $\odot (OHT)$ cuts $L $ at $K\neq 0$, then $\odot (HLK) $, $(OT) $, $(L )$ all pass through $S $. Note, $(OT) $ is the circle of diameter $OT $. Proof of the Lemma- Let $G $ be $\mathrm {I-mixtilinear incircle} $ of $\triangle IBC $, then $(G) $ touches $(L) $, $AB $, $AC $ at $S, X, Y$. Property of Mixtilinear 1. $ST $ passes through $R $, the reflection of point $O $ through $L \implies S\in\odot (OT) $. 2. $SO, BC, XY $ are concurrent at $J $. Now back to the main proof. Let $L'$ be the reflection of $O $ by $H $ and $U $ is the center of $\odot (JHO) $. Let $SR $ cuts $BC $ at $V $. $SR $ is the bisector of $\widehat {BSC} $, and $SJ\perp SR $, $(JBVC)=-1$. Then $HJ\cdot HV=HB^2=HO\cdot HR=HL'\cdot HR\implies V\in\odot (JLR) $. Therefore $SR $, $BC $, $(JLR) $ all pass through $V $. The inversion of center $O $ and $\mathrm {power}(OB^2)$ shows that $(SLH) $, $(L) $, $(OHT) $ all pass through $K $. This implies that $S\in\odot (HLK) $. $(HLK) $, $(L) $ $(OT) $ all pass through $S $. $\blacksquare$

Here is my solution for this problem Solution a) Let $O$ be circumcenter of $\triangle ABC$; $T$ $\equiv$ $AO$ $\cap$ $EF$ Since: $AO$ $\perp$ $EF$ and $A$ is $D$ - excenter of $\triangle DEF$; we have: $FT = FN$, $ET = EN$ Then: $AE$, $AF$ are perpendicular bisectors of $NT$, $TM$ or $N$, $P$, $T$ are collinear and $T$, $Q$, $M$ are collinear So it's easy to see that: $(APQ)$ tangents $EF$ at $T$ b) Let $U$ be midpoint of $MN$; $W$ $\equiv$ $EF$ $\cap$ $MN$ Since: $(MNKW) = (EFTW) = - 1$, we have: $\overline{UK} . \overline{UW} = UN^2 = UM^2 = - \overline{UA} . \overline{UD} = \overline{UL} . \overline{UD}$ or $D$, $L$, $K$, $W$ lie on a circle