a) Let $(I)$ touch $CA$, $AB$ at $U, V$; $I_b, I_c$ be the excenters of vertices $B, C$ in triangle $ABC$. Clearly, $I_aA$, $I_bB$ and $I_cC$ are three altitudes of triangle $I_aI_bI_c$. On the other hand, notice that $IA \perp UV$, $IB \perp VD$, $IC \perp DU$ so two triangles $DUV$ and $I_aI_bI_c$ have corresponding parallel sides. So there exists a homothety $\mathcal{H}$ that turns $D$ into $I_a$, $U$ into $I_b$ and $V$ into $I_c$.

Through $\mathcal{H}$, $I$ becomes $O'$ which is the circumcenter of triangle $I_aI_bI_c$. However, in triangle $I_aI_bI_c$, $I$ is the orthocenter and $(ABC)$ is the Euler circle, so $O$ belongs to $IO'$, which implies the center of $H$ belongs to $OI$. On the other hand, notice that the center of $H$ also lies on $DI_a$; $DI_a$ and $OI$ intersect at $L$ so we infer that $L$ is the center of $H$.

We have $L, E, F$ are collinear and $DE \parallel I_aF$ so $F$ is the image of $E$ via the above homothety, so it lies on $(I_aI_bI_c)$. Furthermore, notice that triangle $I_aI_bI_c$ has $I$ as the orthocenter and $I_aF$ as the altitude, so we infer that $F$ and $I$ are symmetric through $I_cI_b$, hence, $AF = AI$.

b) It is easy to see that $J$ is the midpoint of the minor arc $BC$ of $(O)$, which is a fixed point. Let $K$ be the intersection of $MN$ with the perpendicular bisector of $BC$, we will prove that $K$ is a fixed point, it follows that $T$ always moves on a circle of diameter $JK$, which is a fixed circle. Let $S$ be the midpoint of $BC$, then because $BC$ and $I_cI_b$ are anti-parallel, then $I_aS$ is the symmedian of triangle $I_aI_bI_c$. Since the tangents at $U$ and $V$ of $(I)$ intersect at $A$, $DA$ is the symmedian of the triangle $DUV$. Moreover, the homothety $H$ turns triangle $DUV$ into triangle $I_aI_bI_c$ so we infer that the image of $DA$ through this homothety is $I_aS$. So $I_aS \parallel AD$, infer that $I_a, S$ and $M$ are collinear.

The tangent lines at $B$ and $C$ of $(J)$ intersect at $Q$, we will prove that $K$ is fixed by showing that $K$ is the midpoint of $SQ$.

Let $H$ be the projection of $I_a$ on $BC$. We have $I_aH$ and $I_aJ$ are isogonal with respect to angle $BI_aC$, so it is easy to see $BH = DC$, i.e. $H$ and $D$ are symmetric with respect to $S$. Let $I_aQ$ meet $(J)$ again at $X$ and meet $BC$ at $Y$. We have $I_aQ$ is the symmedian of triangle $I_aBC$, so $X$ and $M$ are symmetric through $BC$, which implies $XH$ passes through $K$.

On the other hand, we have $(QY, XI_a) = -1$ so $H(QY, XI_a) = -1$. Since $QS$ is parallel to $HI_a$ then $K$ is the midpoint of $SQ$. $\square$