11. Given that the internal angles $A, B, C$ of $\triangle ABC$ have opposite sides $a, b, c$ respectively, and $\sqrt{3} b \cos \frac{A+B}{2}=c \sin B$.
(1) Find the size of $\angle C$;
(2) If $a+b=\sqrt{3} c$, find $\sin A$.
(1) $\sqrt{3} \sin B \cos \frac{A+B}{2}=\sin C \sin B \Rightarrow \sqrt{3} \sin \frac{C}{2}=\sin C \Rightarrow \cos \frac{C}{2}=\frac{\sqrt{3}}{2}$,

so $\frac{C}{2}=30^{\circ} \Rightarrow C=60^{\circ}$.
$$
\begin{array}{l}
\text { (2) } \sin A+\sin B=\sqrt{3} \sin C=\frac{3}{2} \Rightarrow \sin A+\sin \left(120^{\circ}-A\right) \\
=\sin A+\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A=\frac{3}{2} \sin A+\frac{\sqrt{3}}{2} \cos A=\frac{3}{2} \\
\Rightarrow \frac{\sqrt{3}}{2} \sin A+\frac{1}{2} \cos A=\sin \left(A+30^{\circ}\right)=\frac{\sqrt{3}}{2} \Rightarrow A=30^{\circ} \text { or } A=90^{\circ} .
\end{array}
$$

So $\sin A=\frac{1}{2}$ or $\sin A=1$.