1) Denote the projections of $I$ on $AC$, $AB$ are $X$, $Y$ respectively.
Note that $B$, $O$, $P$ are collinear because $\angle BMP = 90^\circ$. Because $O$, $H$ are the midpoints of $PB$, $PM$ respectively, we have $OH \parallel BM$ and $OH = \frac{1}{2} BM$.
Similarly, $OK \parallel CN$ and $OK = \frac{1}{2} CN$. Hence, $\angle HOK = \angle EIF$. On the other hand,
$$\begin{aligned} & \frac{OH}{OK} = \frac{BM}{CN} = \frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(A + \frac{C}{2}\right)} \\ & \frac{IF}{IE} = \frac{IY}{\sin \angle IFY} : \frac{IX}{\sin \angle IEX} = \frac{\sin \angle IEX}{\sin \angle IFY} = \frac{\sin \left(A + \frac{B}{2}\right)}{\sin \left(A + \frac{C}{2}\right)} \end{aligned}$$
So $\frac{OH}{OK} = \frac{IF}{IE}$ implies that $\triangle OHK \sim \triangle IFE$.

2) It is easy to see that
$$OK \cdot OT = R^2, \quad OH \cdot OU = R^2$$
with $R$ the radius of $(O)$. Then $OK \cdot OT = OH \cdot OU$ or $K$, $T$, $H$, $O$ are cyclic.
From this, we can see $\angle IEF = \angle IKH = \angle IUT$ so $EF \parallel UT$.
The antipole line of $T$, $U$ passes through $S$ so the antipole line of $S$ is $TU$, which leads to $TU \perp SO$. From these results, we get $SO \perp EF$.