a) Let $T$ be the projection of $A$ on $GD$, it is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\triangle TEB \sim \triangle TFC$ then
$$\frac{TB}{TC} = \frac{BE}{CF}.$$

On the other hand,
$$\frac{BE}{CF} = \frac{BE}{IB} \cdot \frac{IC}{CF} = \frac{\cos ACB}{\cos ABC} = \frac{BD}{CD},$$
which implies that $TBDC$ is a harmonic quadrilateral, or $TD$ passes through $K$, which is the intersection of the tangents at $B, C$ of $(O)$.

b) Let $X, Y$ be the intersections of $KB, KC$ with $IN, IM$, respectively. The tangents at $B, C$ of $(O)$ meet the tangent at $A$ at $X', Y'$. Because $IX \parallel OX'$ and $IY \parallel OY'$, then
$$\triangle KX'Y' \sim \triangle KXY$$
with $O, I$ are corresponding. Note that $O$ is the incenter of $KX'Y'$, it implies that $I$ is the incenter of triangle $KXY$.

Let $(KXY)$ meet $IY, IX$ at $R, S$ respectively. Clearly, $R, S$ are the circumcenters of triangles $KIX$ and $KIY$. Hence, $RS$ is the perpendicular bisector of $IK$ or $H$ lies on $RS$. Applying Pascal's theorem for $\begin{pmatrix} K & X & R \\ S & Y & K \end{pmatrix}$, we obtain that the tangent at $K$ of $(KXY)$, $RS$ and $MN$ are concurrent, which means $KH$ is the tangent of $(KXY)$. By angle chasing, we have
$$\begin{aligned} \angle HIK &= \angle HKI = \angle HKB + \angle BKI \\ &= \angle KYX + 90^\circ - \angle BAC = 270^\circ - 2\angle ABC - \angle BAC \\ &= 90^\circ + \angle ACB - \angle ABC \end{aligned}$$
also, $\angle(AO, BC) = \angle OAC + \angle ACB = 90^\circ - \angle ABC + \angle ACB = \angle HIK$. Note that $IK \perp BC$, thus $IH \perp AO$. Hence, $PBQC$ is cyclic or $I$ has the same power to $(ABC)$ and $(APQ)$, which means $AI$ passes through the second intersection of $(APQ)$ and $(O)$. $\square$