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

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

Let $X$ be the midpoint of arc $BAC$ of $(O)$ then $\triangle XBM = \triangle XCN$. 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 $\triangle ABC$. Since $G$ lies on the circumcircle of $\triangle AMN$, $\triangle AEF$, we have $\triangle GMF \sim \triangle GNE$, 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 $\triangle HBC$. 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,
$$\angle HGQ = 90^\circ - \angle GHE = 90^\circ - \angle CHI \\ = 90^\circ - \angle GPQ = \angle TGQ.$$
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. $\square$