在 $\triangle A B C$ 中, $a, b, c$ 分别是角 $A, B, C$ 的对边, 且 $\frac{\cos B}{\cos C}=-\frac{b}{2 a+c}$.

(1)求角 $B$ 的大小;

(2)若 $b=\sqrt{13}, a+c=4$, 求 $\triangle A B C$ 的面积.
(1) $\because \frac{\cos B}{\cos C}=-\frac{b}{2 a+c}, \therefore$ 由正弦定理得 $\frac{\cos B}{\cos C}=-\frac{\sin B}{2 \sin A+\sin C}$,

即 $2 \sin A \cos B+\sin C \cos B+\cos C \sin B=0$,

$\therefore 2 \sin A \cos B+\sin (B+C)=0$.

$\because B+C=\pi-A, \therefore 2 \sin A \cos B+\sin A=0$.

$\because \sin A \neq 0, \therefore \cos B=-\frac{1}{2}$.

$\because B \in(0, \pi)$,

$\therefore B=\frac{2 \pi}{3}$.

(2) 将 $b=\sqrt{13}, a+c=4, B=\frac{2 \pi}{3}$

代入 $b^{2}=a^{2}+c^{2}-2 a c \cos B$

得 $12=16-2 a c \cdot\left(1-\frac{1}{2}\right)$,

$\therefore a c=3$,

$\therefore S_{\triangle A B C}=\frac{1}{2} a c \sin B=\frac{1}{2} \times 3 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4}$.