Since $CD$ is diameter of $ \odot(CMN),$ then $O$ is circumcenter of $\triangle CMN.$ Further $\angle CMN=\angle CDN=\angle BAC=60^{\circ}$ $\Longrightarrow$ $\angle CON=\angle CHN=120^{\circ}$ $\Longrightarrow$ $CHON$ is cyclic $\Longrightarrow$ $\angle COH=\angle CNH=90^{\circ}-75^{\circ}=15^{\circ}.$

$\overline{NO}$ is a median of 30-60-90 $\triangle CND$, so $\angle CNO = 30^{\circ}$. $\angle CNM = \angle CDM = 45^{\circ}$. Thus $\angle CNH = \angle HNO = \angle MNO = 15^{\circ}$. $\angle HCM = \angle HNM = 30^{\circ}$, so $\angle OCH = 15^{\circ}$. So $CHON$ is cyclic and $\angle COH = 15^{\circ}$.