13. In $\triangle A B C$, $\angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. Let
$$
\begin{array}{l}
f(x)=\boldsymbol{m} \cdot \boldsymbol{n}, \boldsymbol{m}=(2 \cos x, 1), \\
\boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x), \\
f(A)=2, b=1, S_{\triangle A B C}=\frac{\sqrt{3}}{2} . \\
\text { Then } \frac{b+\boldsymbol{c}}{\sin B+\sin C}=
\end{array}
$$
II. 13.2.
It is known that $f(x)=1+2 \sin \left(2 x+\frac{\pi}{6}\right)$.
From the given conditions, we have $\angle A=\frac{\pi}{3}, c=2, a=\sqrt{3}$. Therefore, $\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=2$.