13. In $\triangle A B C$, $a, b, c$ are the sides opposite to angles $A, B, C$ respectively, $b=1$, and $\cos C+(2 a+c) \cos B=0$.
(1) Find $B$;
(2) Find the maximum area of $\triangle A B C$.
(1) $b \cos C+(2 a+c) \cos B=0 \Rightarrow \sin A+2 \sin A \cos B=0 \Rightarrow \cos B=-\frac{1}{2} \Rightarrow B=120^{\circ}$.
(2) $1=b^{2}=a^{2}+c^{2}+a c \geqslant 3 a c \Rightarrow a c \leqslant \frac{1}{3}$, equality holds when $a=c=\frac{\sqrt{3}}{3}$. Therefore, the maximum area of $\triangle ABC$ is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{12}$.