My solution. Let $O$ be the circumcenter of $\odot(ABC)$, $P\equiv AO\cap DE$, $X\equiv AO\cap BC$, $L$ be the midpoint of $EF$. (a). I'll claim that $D \in \odot(AMN)$. Note that $\angle EDB=90^\circ-\angle ABC=\angle OAC \implies P,D,C,A$ are concyclic $\implies \angle DEF=\angle DAF=\angle DPC \implies EF\parallel CP$. Thus $\angle MND=\angle PCD=\angle PAD$ as desired. (b). I'll cliam that $T,O,D$ are collinear. At the beginning, see CGMO P1 here. CGMO P1 give us $DL\parallel AO \implies \angle LDE=\angle APE=90^\circ - \angle PAE=\angle ACB=\angle ODF \implies OD$ is symmedian wrt $\odot(AFDE)$. The result then follows. 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My solution for part b) Let $DO\cap \odot (AEF)=X$, then $(X,D;E,F)=A(\infty,D;B,C)=-1\implies XEDF$ is harmonic quadrilateral. Hence $T,D,X$ collinear so $T$ lies on perpendicular bisector of $BC$(which is fixed).

Here is my solution for this problem a) We have: ($DA$; $DB$) $\equiv$ ($DA$; $DE$) + ($DE$; $DB$) $\equiv$ ($FA$; $FE$) + $\dfrac{\pi}{2}$ $-$ ($BC$; $BA$) $\equiv$ ($FA$; $FE$) + ($AO$; $AC$) $\equiv$ ($NA$; $NM$) (mod $\pi$) So: $D$ $\in$ ($AMN$) b) Let $S$ $\equiv$ $DO$ $\cap$ ($AEF$) ($S$ $\not \equiv$ $D$) then $AS$ $\parallel$ $BC$ We have: $\dfrac{SE}{SF}$ = $\dfrac{AD . \sin \widehat{SFE}}{AD . \sin \widehat{SEF}}$ = $\dfrac{\sin \widehat{SAE}}{\sin \widehat{FAS}}$ = $\dfrac{\sin \widehat{ABC}}{\sin \widehat{ACB}}$ = $\dfrac{DE}{BD}$ . $\dfrac{DC}{DF}$ = $\dfrac{DE}{DF}$ So: $DESF$ is harmonic quadrilateral Hence: tangents at $E$, $F$ of ($AEF$) and $DS$ concurrent at $T$ or $T$ lies on perpendicular bisector of $BC$