1) It is easy to check that $BQ$ is the diameter of circle $(O)$. Denote $I$ as the intersection of $SR$ and $PK$. We have
$$\angle SPI = \angle SBK = \angle QCA$$
and
$$\angle PSI = \angle PBR = \angle CQR.$$
This implies that two triangles $PSI$ and $CQM$ are similar. By the same way, we also have two triangles $KSI$ and $AQM$ are similar. But $M$ is the midpoint of the segment $AC$ then $I$ is the midpoint of $PK$.

2) Redefine point $E$ as the projection of $C$ to $AB$. Denote $H$ as the intersection of the altitudes $AD$, $BL$ of triangle $ABC$. Note that the quadrilateral $LMND$ is cyclic then $CM \cdot CL = CN \cdot CD$ and $HA \cdot HD = HB \cdot HL$, which implies that $CH$ is the radical axis of two circles of diameter $AN$ and $BM$. We have $E$ belongs to $CH$ then it also belongs to the radical axis of two circles of diameter $AN$ and $BM$. Since $\angle BEH = \angle BRP = 90^\circ$ then the quadrilateral $BHER$ is cyclic, thus
$$\angle HRE = \angle EBH = \angle OBC = \angle PBS = \angle PRS$$
which implies that $S$, $R$, $E$ are collinear. Then $E$ is the intersection of $SR$ and the radical axis of two circles of diameter $AN$, $BM$. Denote $X$ as the intersection of $EM$ and $BQ$. Let $T$ be the midpoint of segment $BC$ then the quadrilateral $ETML$ is cyclic. Thus
$$\angle MEC = 90^\circ - \angle MET = 90^\circ - \angle ALT = 90^\circ - \angle BAC = \angle BAL = \angle BCQ.$$
This means that the quadrilateral $BCXE$ is cyclic, and then $\angle BXC = BEC = 90^\circ$. Therefore, $X$ is the projection of $C$ onto $BQ$, which is a fixed point and the line $EM$ passes through $X$.