For $(b)$ let $D$ be the midpoint of $BC$ and $G=AC \cap XN$. Noting $\triangle BMC \cap D \simeq \triangle MXC \cap N$ and $MD \parallel AC$ we get: $$\measuredangle GXM=\measuredangle NXM=\measuredangle DMB=\measuredangle CAB=\measuredangle GAM$$So $AGXM$ are concyclic as desired.

For $(a)$, Let $CX \cap \omega=F \neq C$. Let the feet of the perpendiculars from $F,A$ to $CM$ be $K,L$ respectively. $$\measuredangle AFC=\measuredangle AFX=\measuredangle AMX=\measuredangle AMC-\measuredangle XMC=\measuredangle CXM-\measuredangle XMC=\measuredangle MCX=\measuredangle MCF$$So $AF \parallel MC$ $$CF=\frac{FK}{\sin{\angle FCM}}=\frac{AL}{\sin{\angle MCB}}=\frac{\frac{[ABC]}{MC}}{\sin{\angle MCB}}=\frac{BC \cdot CM \cdot \sin{\angle MCB}}{MC \sin{\angle MCB}}=BC$$Also by $\triangle BMC \cap D \simeq \triangle MXC \cap N$: $$CX=\frac{CM^2}{BC}$$So: $$CX \cdot CF=\frac{CM^2}{BC} \cdot BC=CM^2$$Hence $CM$ is tangent to the circle

sbealing wrote: For $(a)$, Let $CX \cap \omega=F \neq C$. Let the feet of the perpendiculars from $F,A$ to $CM$ be $K,L$ respectively. $$\measuredangle AFC=\measuredangle AFX=\measuredangle AMX=\measuredangle AMC-\measuredangle XMC=\measuredangle CXM-\measuredangle XMC=\measuredangle MCX=\measuredangle MCF$$So $AF \parallel MC$ $$CF=\frac{FK}{\sin{\angle FCM}}=\frac{AL}{\sin{\angle MCB}}=\frac{\frac{[ABC]}{MC}}{\sin{\angle MCB}}=\frac{BC \cdot CM \cdot \sin{\angle MCB}}{MC \sin{\angle MCB}}=BC$$Also by $\triangle BMC \cap D \simeq \triangle MXC \cap N$: $$CX=\frac{CM^2}{BC}$$So: $$CX \cdot CF=\frac{CM^2}{BC} \cdot BC=CM^2$$Hence $CM$ is tangent to the circle Good job

$\textbf{(a)}$ Let, $CX \cap \omega \equiv Y$ and $CM \cap BY \equiv Z$ $\color{green} \textbf{Claim 1 :}$ $BC=CY$ $\textbf{Proof :}$ $$\angle ZMA= \angle BMC= \angle MXC= 180 -\angle YAM \implies AY \parallel CM\equiv CZ$$But $M$ is the midpoint of $AB$ and $MZ \parallel AY \implies Z$ is the midpoint of $BY$ Also we have $\angle BCZ =\angle ZCY \implies BC=CY \square$ $\color{green} \textbf{Claim 2 :}$ $CM^2= CX.CY$ $\textbf{Proof :}$ We have $$\triangle BCM \sim \triangle MCX \implies CM^2=CX.BC=CX.CY\square$$ Hence $CM$ is tangent to $\omega \blacksquare$ $\textbf{(b)}$ Let $P$ be midpoint of $BC$ and $AC \cap \omega \equiv J$ then $\angle MAJ= \angle BMP =\angle MXN \implies N,X,J$ are collinear $\blacksquare$