11. In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and $a^{2}-(b-c)^{2}=(2-\sqrt{3}) b c$, $\sin A \sin B=\cos ^{2} \frac{C}{2}$, the length of the median $A M$ on side $B C$ is $\sqrt{7}$.
(1) Find the sizes of $A$ and $B$;
(2) Find the area of $\triangle A B C$.
$$
\begin{array}{l}
\text { (1) } a^{2}-b^{2}-c^{2}=-\sqrt{3} b c \Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{\sqrt{3}}{2} \Rightarrow A=30^{\circ} \text {; } \\
\sin A \sin B=\frac{1+\cos C}{2}=\frac{1}{2}-\frac{1}{2}(\cos A \cos B-\sin A \sin B) \\
\Rightarrow \cos A \cos B+\sin A \sin B=\cos (A-B)=1 \Rightarrow A=B=30^{\circ} .
\end{array}
$$
(2) From part (1), we know that $\triangle A B C$ is an isosceles triangle with the vertex angle of $120^{\circ}$, let $A B=\sqrt{3} x$
$$
\Rightarrow A M^{2}=3 x^{2}+\frac{1}{4} x^{2}-\sqrt{3} x^{2} \cdot \frac{\sqrt{3}}{2}=\frac{7}{4} x^{2}=7 \Rightarrow x=2 \text {. }
$$

Therefore, $S_{\triangle A B C}=\frac{\sqrt{3}}{4} x^{2}=\sqrt{3}$.