a) Let $T$ be the foot of $A$ on $EF$. Note that $FC$ is the internal bisector of $\angle DFE$ so $FM$, $FT$ are symmetric with respect to $AB$. On the other hand, $\angle FNA = \angle FTA = 90^\circ$ then $N$, $T$ are symmetric with respect to $AB$. Therefore, $N$, $P$, $T$ are collinear and $TP \perp AB$. Similarly, $TQ \perp AC$ and $T$, $M$, $Q$ are collinear. Hence, $AT$ is the diameter of $(I)$ and $AT \perp EF$ then $EF$ is tangent to $(I)$.

b) Let $X$, $Y$, $Z$ and $J$ be the intersections of $MN$ with $EF$, $AB$, $AC$ and $AD$. It is well-known that $M$, $N$ are the reflections of $T$ through $AE$, $AF$ respectively then $MN$ passes through the feet of altitudes from $E$, $F$ in triangle $AEF$. Thus, $EY \perp AB$ and $FZ \perp AC$. On the other hand, we also have $AD$ is the perpendicular bisector of $MN$ then $J$ is the midpoint of $MN$. Hence, $D(EF, TX) = -1$. Projecting on $MN$, we have $(MN, KX) = -1$. Note that $J$ is the midpoint of $MN$, we have
$$\overline{JK} \cdot \overline{JX} = JM^2 = JN^2 = -\overline{JA} \cdot \overline{JD} = \overline{JL} \cdot \overline{JD}.$$
Therefore, $DLKS$ is cyclic or $(DLK)$ passes through $S$. $\square$