a) Since $D, E$ are the tangency points of $(I)$ with $BC$, $CA$ respectively, we have $DE \perp CI$. Moreover, $FL \perp CI$ so $DE \parallel FL$. Similarly, we have $DF \parallel EK$.

Clearly, we have $DK = DL = EF$ or $D$ lies on the perpendicular bisector of $KL$. Furthermore, $I$ also lies on the perpendicular bisector of $KL$, therefore $D, I$ and $J$ are collinear.

b) Let $T$ be the midpoint of $BC$, $G$ be the intersection of $AI$ with $BC$ and $U$ be the reflection of $T$ over $G$. We will show that the perpendicular bisector of $PQ$ always passes through $U$. Let $X, Y$ be the intersections of $EF$ with $BI$ and $CI$, respectively. We have
$$\angle XIC = 180^\circ - \angle BIC = 90^\circ - \frac{\angle BAC}{2} \\ = \angle AEI - \angle EFI = \angle AEF = \angle XEC.$$
Hence, $C, I, E$ and $X$ are concyclic, which implies $\angle BXC = 90^\circ$. Similarly, we obtain $\angle BYC = 90^\circ$. Since $\angle BXC = \angle BYC = 90^\circ$, the four points $B, C, X$ and $Y$ lie on a circle with center $T$ and diameter $BC$. It follows that $T$ lies on the perpendicular bisector of the segment $XY$.

It is easy to see that the pairs of points $M, E$ and $N, F$ are symmetric with respect to $I$, so $EF \parallel MN$ or $XY \parallel PQ$. From there, we have $\triangle IEX = \triangle IMP$, which implies that $X$ and $P$ are symmetric with respect to $I$. Similarly, we also have $Y$ and $Q$ are symmetric with respect to $I$. Hence two perpendicular bisectors of $XY$ and $PQ$ are symmetric with respect to $AI$. Thus, the perpendicular bisector of $PQ$ passes through $U$.

Since $B, C$ are fixed, then $T$ is fixed. Note that $\frac{GB}{GC} = \frac{AB}{AC}$ remains unchanged so $G$ is fixed. Hence $U$ is fixed. So the perpendicular bisector of $PQ$ always passes through the fixed point $U$. $\square$