1) Denote $J$ as the center of the nine-point circle of triangle $ABC$, then $(J)$ passes through $E$, $I$, $F$ and point $J$ is also the midpoint of segment $OH$. It is easy to see that $D$ and $H$ are symmetric with respect to the line $BC$, then triangle $BDH$ is isosceles with $BD = BH$. Since triangle $IEH$ has $IE = IH$ then

$$\angle IEH = \angle IHE = \angle BHD = \angle BDH,$$
which implies that $BDEI$ is a cyclic quadrilateral. But $DB$ cuts $EI$ at $M$ then
$$\overline{ME} \cdot \overline{MI} = \overline{MB} \cdot \overline{MD}.$$

Thus the power of point $M$ to circles $(J)$ and $(O)$ are equal. Similarly, the power of point $N$ to circles $(J)$ and $(O)$ are also equal. So we can conclude that $MN$ is the radical axis of $(O)$ and $(J)$, thus $MN \perp OJ$. But $O$, $H$, $J$ are collinear then $MN \perp OH$.

2) Let $X$ be the midpoint of $EF$ and $K$ be the intersection of $AH$ and $BC$. It is easy to see that two triangles $BFE$ and $KHE$ are similar, which implies that two triangles $BFX$ and $DHE$ are also similar, thus $\angle FBX = \angle HDE = \angle FBP$. Then three points $B$, $X$, $P$ are collinear; similar to three points $C$, $X$, $Q$.

Denote $AL$ as the diameter of circle $(O)$ then we can see that $SH$ passes through $L$ and quadrilateral $HBLC$ is a parallelogram, which implies that $HL$ passes through the midpoint $M$ of $BC$. It is easy to check that two triangles $SEC$ and $SFB$ are similar then two triangles $SEF$ and $SCB$ are also similar. These triangles have the medians $SX$ and $SM$ respectively then $\angle FSX = \angle BSM$. We also have two triangles $SFB$ and $SRL$ are similar then two triangles $SFR$ and $SBL$ are also similar. Thus
$$\angle FSR = \angle BSL = \angle BSM = \angle FSX.$$
From this we can conclude that three points $S$, $X$, $R$ are collinear or $SR$ passes through $X$. Therefore, three lines $BP$, $CQ$ and $RS$ are concurrent at the midpoint $X$ of the segment $EF$.