a) Let $T$ be the second intersection of $(BDE)$ and $(CDF)$. We have $BF = CE$ and
$$\angle TED = \angle TBD, \quad \angle TCD = \angle TFD$$
so two triangles $TCE$ and $TFB$ are congruent. From this it is easy to obtain that $TBE$ and $TCF$ are isosceles triangles at $T$ and they are similar.

It implies that $T$, $I$ and $M$ are collinear; $T$, $J$ and $N$ are collinear. Note that $\frac{TI}{TM} = \frac{TJ}{TN}$ so $MN$ is parallel to $IJ$.

b) Let $X$, $Y$ be the midpoints of $CE$, $BF$ and $\ell$ be the radical axis of the circles with diameter $BF$ and $CE$. Two altitudes $BX'$ and $CY'$ of triangle $ABC$ intersect at orthocenter $S$.

We will prove that $HK$ passes through a fixed point $S$ by showing that these three points lie on $\ell$. Indeed, we see that $MX$ is the midline of triangle $EBC$, so $MX \parallel BC$, similarly, we have $NY \parallel BC$ so $MX \parallel NY$. Similarly, we also have $MY \parallel NX$ so $MYNX$ is a parallelogram, so $K$ is the midpoint of $XY$. From there, notice that $CE = BF$ then
$$\mathcal{P}_{K/(BF)} = KY^2 - BY^2 = KX^2 - CX^2 = \mathcal{P}_{K/(CE)},$$
which implies $K \in \ell$.

Besides, we have $BCX'Y'$ is a cyclic quadrilateral so
$$\mathcal{P}_{S/(BF)} = \overline{SX'} \cdot \overline{SB} = \overline{SY'} \cdot \overline{SC} = \mathcal{P}_{S/(CE)},$$
hence $S \in \ell$. Similarly, we can point out that $H$ lies on $\ell$, thus $HK$ passes through a fixed point $S$.