13. Given $f(x)=\boldsymbol{m} \cdot \boldsymbol{n}$, where $\boldsymbol{m}=(2 \cos x, 1), \boldsymbol{n}=(\cos x, \sqrt{3} \sin 2 x)$. In $\triangle A B C$, $\angle A$ satisfies $f(A)=2$, $\angle B$ is opposite to side $b=1$, and the area of $\triangle A B C$ is $S=\frac{\sqrt{3}}{2}$. Then $\frac{b+c}{\sin B+\sin C}=$ $\qquad$ .
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x+1=2 \sin \left(2 x+\frac{\pi}{6}\right)+1 \Rightarrow \sin \left(2 A+\frac{\pi}{6}\right)=\frac{1}{2} \text {, }
$$

Thus $A=\frac{\pi}{3} \Rightarrow b c=2 \Rightarrow c=2$. Then $a^{2}=b^{2}+c^{2}-b c=3 \Rightarrow 2 R=\frac{a}{\sin A}=2 \Rightarrow \frac{b+c}{\sin B+\sin C}=2$.