a) The midpoint $E$ of arc $AB$ where $C$ does not lie, lies on the bisector $CI$. Similarly, $F$ lies on $BI$. It holds that $\angle IFC = \angle BFC = \angle BAC = 90^{\circ}$ because $ABCF$ is a cyclic quadrilateral and $\angle IDC = 90^{\circ}$ because $D$ is the tangency point of the inscribed circle with $BC$. Therefore, $\angle IFC + \angle IDC = 180^{\circ}$, which implies that $FIDC$ is a cyclic quadrilateral. Now, $\angle DFI = \angle DCI = \frac{1}{2} \angle ACB$, while also $\angle IFE = \angle BFE = \angle BCE = \frac{1}{2} \angle ACB$. Thus, $\angle DFE = \frac{1}{2} \angle ACB + \frac{1}{2} \angle ACB = \angle ACB$. Similarly, $\angle DEF = \angle ABC$. Together, this gives $\triangle DEF \sim \triangle ABC$.
b) Let $S$ be the intersection of $EF$ with $AB$. In the previous part, we have seen that $BF$ is the bisector of $\angle DFE = \angle DFS$. Also, $BF$ is the bisector of $\angle ABC = \angle SBD$. Therefore, $\triangle BDF \cong \triangle BSF$ by (SAS). This means that $|BD| = |BS|$. On the other hand, the distances from $B$ to the tangency points of the inscribed circle with $BC$ and $BA$ are equal, and one of these tangency points is $D$, so the other tangency point must be $S$. Thus, $EF$ passes through the tangency point of the inscribed circle with $AB$. Similarly, it also passes through the tangency point of the inscribed circle with $AC$.