Hint

A solution for the first problem: Let $S= AB \cap (FPQ)$, $T= AC \cap (EPQ)$, $M$ the midpoint of $BC$ and $N = AM \cap ST$. Then we have that: $AF \cdot AS = AP \cdot AQ = AE \cdot AT$ and hence $SFET$ is cyclic. So: $\angle AST = \angle AEF = \angle ABC => ST \parallel BC => N$ is the midpoint of $ST => NS=NT (1)$. Also from the cyclic quadrilaterals we get that: $\angle SPQ = \angle SFQ = \angle BFC =90 = \angle BEC = \angle PET = \angle MQT$, hence $SP \parallel QT (2)$, as they are both perpendicular to $AM$. From $(1)$ and $(2)$ we get that $SPTQ$ is a parallelogram. Using the parallelogram and the cyclic quadrilaterals: $\angle AFP = \angle PQS = \angle QPT = \angle QET$. Now let $X$ be a point such that $XC<XB$ and $AX$ is tangent to $(AQE)$. Then we have that: $\angle AFP = \angle QET = \angle QAE + \angle AQE = \angle QAE + \angle XAE = \angle XAQ = \angle XAP$, hence $AX$ is also tangent to $(APF)$, which proves the desired tangency.

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For the second problem: Let $O_1$ be the center of $(AFP)$, $O_2$ be the center of $(AQE)$ and $H= BE \cap CF$ . By radical axis we have that $O_1O_2 \perp AR$ so it suffices to prove that $O_1O_2 \parallel \ell$. Since $O_1$ is the circumcenter of $AFP$ we conclude that: $\angle FO_1P = 2 \angle FAP = 2\angle BAP = 2 \angle A/2 = \angle A = 180 - \angle BHC = 180 - \angle FHP$, hence $FHPO_1$ is cyclic. Hence, using that $FO_1=PO_1$, we have that $O_1$ is the midpoint of the arc $FP$ (that does not contain $H$) in $(FHP)$. So $HO_1$ is angle bisector of $\angle FHP$, and hence angle bisector of $\angle BHC$, which is obviously parallel to $\ell$. Hence $HO_1 \parallel \ell$ and analogously we have that $HO_2 \parallel \ell$. So we have that $O_1O_2 \parallel \ell$, as needed.

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For $\mathrm{i)}$ Let $M$be the mid-point of $BC$,suppose $X,Y\in AM$ satisfy $\angle BXM=90^{\circ}-\angle ACB,\angle CYM=90^{\circ}-\angle ABC$ respectively, $T$ be the $H$-Humpty point of $\triangle ABC$ Consider an inversion $\mathcal{I}^{M}_{MB^{2}}$ So $P\mapsto X,Q\mapsto Y,A\mapsto T,F\mapsto F,E\mapsto E$ Hence $\odot(APF)\mapsto \odot(XFT),\odot(AQE)\mapsto \odot(YET)$ We need to prove that $\odot(XFT)$ is tangent to $\odot(YET)\Rightarrow \angle YET+\angle XFT=180^{\circ}$ Note that $H,B,C,T$ are cyclic Hence $\angle HTC=180^{\circ}-\angle HBC=180^{\circ}-\angle BXM=\angle AXB$ Note that $\angle TBM=\angle BAM$ Hence $\angle THC=\angle BAX$ Which means that $\triangle BAX\sim \triangle CTH$ Similarly $\triangle CYA\sim \triangle BTH$ Consider that $MP\cdot MX=MB^{2}=ME^{2}$ Hence $\angle EXM=\angle BEM=\angle EBM$ Which means that $\angle AXE=180^{\circ}-\angle TXE=180^{\circ}-\angle EBC=180^{\circ}-\angle EAH=\angle HTE$ So $\triangle EAX\sim \triangle EHT$ Similarly $\triangle FYA\sim \triangle FTH$ Since $\dfrac{XB}{YC}=\dfrac{\dfrac{AX\cdot TC}{HT}}{\dfrac{AY\cdot BT}{HT}}=\dfrac{AX}{AY}\cdot \dfrac{TC}{TB}=\dfrac{\dfrac{AX}{HT}}{\dfrac{AY}{HT}}\cdot \dfrac{AC}{AB}=\dfrac{AE}{AF}\cdot \dfrac{HF}{HE}\cdot \dfrac{AC}{AB}=\dfrac{HF}{HE}$ Which means that $\triangle XBF\sim \triangle YCE$ So $\angle YET+\angle XFT=\angle AET+180^{\circ}-\angle YEC+\angle AFT-180^{\circ}+\angle TFB=\angle AET+\angle AFT$ Note that $A,F,H,T,E$ are cyclic So $\angle YET+\angle XFT=\angle AET+\angle AFT=180^{\circ}$ Which means that $\odot(APF)$ and $\odot(AQE)$ are tangent

For $\mathrm{ii)}$ Let $O_{1},O_{2}$ be the circumcenter of $\triangle AFP$ and $\triangle AEQ$ respectively Note that $AR$ is the root axis of $\odot(AFP)$ and $\odot(AEQ)$ so we need to prove that $O_{1}O_{2}//AQ$ Consider that $\angle FO_{1}P=\angle BAC=180^{\circ}-\angle FHP$ Which means that $F,O_{1},P,H$ are cyclic Hence $\angle O_{1}HF=\angle O_{1}PF=90^{\circ}-\dfrac{1}{2}\angle BAC$ Similarly $\angle O_{2}HQ=180^{\circ}-\angle O_{2}EQ=90^{\circ}+\dfrac{1}{2}\angle BAC$ So $\angle O_{1}HF+\angle O_{2}HF=180^{\circ}$ Which means that $O_{1},O_{2},H$ are collinear Since $\angle O_{1}HF=\angle O_{1}PF=90^{\circ}-\angle FAQ=\angle AQF$ Hence $O_{1}O_{2}//AQ$ Which means that $AR\bot l$