In case $AB = AC$, it is obvious that $DB = DC$ and the bisector of $\angle MTN$ passes through the midpoint of the minor arc $BC$ of $O$ (which is the fixed point that is revealed in part b), hence we only need to consider the case where $AB \neq AC$.

a) Let $R, S$ be the second intersections of $I$ with $AB, AC$ respectively. Since $E, F$ lie on a circle with diameter $BC$, and quadrilateral $EFRS$ is cyclic, we conclude $BC \parallel RS$ by Reim's theorem.

By the assumption, $I$ touches $BC$ at $D$ then
$$\frac{BD^2}{CD^2} = \frac{BF \cdot BR}{CE \cdot CS} = \frac{BF \cdot BR}{CE \cdot CS} = \frac{BF \cdot AB}{CE \cdot AC} = \frac{BF \cdot BE}{CE \cdot CF} = \frac{\cot B}{\cot C}$$
which means $\frac{BD}{CD} = \sqrt{\frac{\cot B}{\cot C}}$.

b) Denoted by $G$ the intersection of $EF$ and $BC$. We have
* The radical axis of $(BHC)$ and $I$ is $PQ$,
* The radical axis of $I$ and $(BFCE)$ is $EF$,
* The radical axis of $(BHC)$ and $(BFCE)$ is $BC$.

Hence $G$ is the radical center of these circles, which means $G$ lies on $PQ$. On the other hand, we have
* Let the radical axis of $O$ and $(TPQ)$ be $d$,
* The radical axis of $(TPQ)$ and $(BHC)$ is $PQ$,
* The radical axis of $(BHC)$ and $O$ is $BC$.

then $d$, $PQ$ and $BC$ concur at $G$ which means $GT$ is the radical axis of $O$ and $(TPQ)$, hence $GT$ touches $(TPQ)$ and $O$. Since $PQMN$ is a cyclic quadrilateral,
$$\overline{GM} \cdot \overline{GN} = \overline{GP} \cdot \overline{GQ} = GT^2,$$
hence $GT$ touches $(TMN)$.

In conclusion, $(TMN)$ touches $O$ at $T$, we conclude that $TM, TN$ are isogonal with respect to $\angle BTC$ then the bisector of $\angle MTN$ is coincident with the bisector of $\angle BTC$, which passes through $J$, the midpoint of minor arc $BC$ of $O$. $\square$