a/ First, we show $MN \parallel OA$.
Indeed, without loss of generality assume that $\overline{ABC} > \overline{ACB}$. Then
$$\overline{OAD} = \overline{OAB} - \overline{DAB} = \frac{180^\circ - \overline{AOB}}{2} - \frac{\overline{BAC}}{2} = 90^\circ - \left( \hat{C} + \frac{\overline{BAC}}{2} \right) \quad (1)$$
Since $AD$ and $AE$ are internal and external bisectors of $BAC$, we have $AD \perp AE$. Together
with the assumption that $I$ is the midpoint of $DE$, we conclude
$$\overline{IAE} = \overline{IEA} = 90^\circ - \overline{ADE} = 90^\circ - \left( \frac{\overline{BAC}}{2} + \hat{C} \right) \quad (2)$$
From (1) and (2) it follows that $\overline{OAD} = \overline{IAE}$.

Hence $\overline{OAI} = \overline{OAD} + \overline{DAI} = \overline{IAE} + \overline{DAI} = 90^\circ$, or $OA \perp AI$. But $MN \perp AI$ (assumption)
hence $MN \parallel OA$. (3)
Let $H$ be the orthocenter of triangle $ABC$ and $K$ be the midpoint of $BC$, we have $H \in MN$ (assumption) and $\overline{AH} = 2\overline{OK}$. Since $O$ and $K$ are fixed, the vector $\overline{v} = 2\overline{OK}$ is constant.
Together with (3), this implies $MN$ is the image of $OA$ under the translation along vector $\overline{v}$.
Hence, let $O'$ be the image of $O$ under the translation along vector $\overline{v}$, we have $O' \in MN$.
Since $O$ is fixed, $O'$ is also fixed. Thus $MN$ always passes through a fixed point.

b/ It follows from (3) that $\overline{AMN} = \overline{OAD}$. Together with (1) and (2) this implies $\overline{AMN} = \overline{IEA}$.
But $\overline{MAH} = \overline{IEA}$ (angles with parallel sides) hence $\overline{AMN} = \overline{MAH}$. Consequently, since $AMN$ is a right-angled triangle in $A$, $H$ is the midpoint of $MN$. Thus $MN = 2AH = 4OK = \text{const}$. Hence, $AMN$ has the largest area if and only if it is isosceles in $A$.
We have: $\triangle AMN$ isosceles and right-angled in $A \Leftrightarrow MN \perp AH \Leftrightarrow MN \parallel BC \Leftrightarrow OA \parallel BC$.
Thus, the triangle $AMN$ has the largest area if and only if $A$ is in the two ends of the diameter parallel with $BC$.