TST

a) It is clear that $G$ is the Miquel point of completed quadrilateral $MNEF.LA$, thus $G$ lies on the circumcircle of $△AMN$, $△AEF$. Besides,

$$BM=BF+FM=EN+CE=CN.$$



Let $X$ be the midpoint of arc $BAC$ of $(O)$ then $△XBM=△XCN$. Hence, $\angle XMA=\angle XNA$; it follows that $X$ lies on the circumcircle of $AMN$, in other words, $(GMN)$ passes through fixed point $X$.

b) Let $H$ be the orthocenter of $△ABC$. Since $G$ lies on the circumcircle of $△AMN$, $△AEF$, we have $△GMF\sim △GNE$, thus $$\frac{GE}{GF}=\frac{NE}{MF}=\frac{BF}{CE}=\frac{HF}{HE}$$ This means $GH$ bisects $EF$. Since $EF$ and $BC$ are anti-parallel with respect to $\angle BHC$, therefore $HG$ is the symmedian of $△HBC$. It is well known that $K$ and $H$ are symmetric with respect to $BC$. Thus $\angle PDI=\angle DKI=\angle DHI$, we obtain that $PQ\perp HI$. Therefore, $$\begin{gathered} \angle HGQ=90^{\circ }-\angle GHE=90^{\circ }-\angle CHI \\ =90^{\circ }-\angle GPQ=\angle TGQ. \end{gathered}$$ Hence, $G$, $H$ and $T$ are collinear. It is clear that $(BHC)$ is fixed circle since $\angle BHC=180^{\circ }-\angle BAC$. Thus, $HT$ passes through the intersection of tangents at $B$, $C$ of $(BHC)$ which is fixed. $\square$