1) From the power of point to circle, we have
$$\mathscr{P}_{A /(BIF)} = AF \cdot AB \text{ and } \mathscr{P}_{A /(CIE)} = AE \cdot AC.$$
But it is easy to see that $AF \cdot AB = AE \cdot AC$ then $A$ belongs to the radical axis of two circles $(BIF)$, $(CIE)$.
Hence, $A$, $D$, $I$ are collinear.
Denote $J = AH \cap BC$ then $AD \cdot AI = AH \cdot AJ$ which implies that $H$, $D$, $I$, $J$ are concyclic, then $HD \perp AI$. Then by considering the radical axis of three circles $(AEF)$, $(JFEI)$, $(JHDI)$, we can conclude that $BC$, $HD$, $EF$ are concurrent.

2) Using sine law in some triangle, we get
$$\frac{DB}{\sin \angle BID} = \frac{BI}{\sin \angle BDI} = \frac{BC}{2 \sin \angle ABC}$$
and
$$\frac{DC}{\sin \angle CID} = \frac{CI}{\sin \angle CDI} = \frac{BC}{2 \sin \angle ACB}$$
Then
$$\frac{DB}{DC} = \frac{\sin \angle ACB}{\sin \angle ABC} = \frac{AB}{AC} = \frac{LB}{LC}.$$
This implies that $(ADL)$ is the Apollonius circle of triangle $ABC$.
Thus $\frac{KB}{KC} = \frac{AB}{AC}$ or $ABKC$ is the harmonic quadrilateral, then $AK$ is the symmedian of triangle $ABC$. Hence, $AI$, $AK$ are isogonal conjugate in angle $\angle BAC$, also in angle $\angle EAF$. Since $G$, $D \in (AEF)$, we have $GD \parallel EF$.
Finally, $AS$ is the external bisector of the angle $\angle BAC$, also of the angle $\angle EAF$ then $T$ is the midpoint of major $\operatorname{arc} EF$. From these, we can conclude that $TD = TG$. $\square$