13. Given the function $f(x)=4 \cos x \cdot \sin \left(x+\frac{7 \pi}{6}\right)+a$ has a maximum value of 2.
(1) Find the value of $a$ and the smallest positive period of $f(x)$; (2) Find the intervals where $f(x)$ is monotonically decreasing.
(1)
$$
\begin{aligned}
f(x) & =4 \cos x \cdot \sin \left(x+\frac{7 \pi}{6}\right)+a=4 \cos x \cdot\left(-\frac{\sqrt{3}}{2} \sin x-\frac{1}{2} \cos x\right)+a \\
& =-2 \sqrt{3} \sin x \cos x-2 \cos ^{2} x+1-1+a=-\sqrt{3} \sin 2 x-\cos 2 x-1+a \\
& =-2 \sin \left(2 x+\frac{\pi}{6}\right)-1+a
\end{aligned}
$$

Therefore, when $\sin \left(2 x+\frac{\pi}{6}\right)=-1$, $f(x)$ reaches its maximum value $2-1+a=1+a$.
Since the maximum value of $f(x)$ is 2, we have $1+a=2$, which means $a=1$.
The smallest positive period of $f(x)$ is $T=\frac{2 \pi}{2}=\pi$;
(2) From (1), we get $f(x)=-2 \sin \left(2 x+\frac{\pi}{6}\right)$.
Let $2 x+\frac{\pi}{6} \in\left[-\frac{\pi}{2}+2 k \pi, \frac{\pi}{2}+2 k \pi\right], k \in \mathbf{Z}$, we get $x \in\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right], k \in \mathbf{Z}$.
Therefore, the decreasing interval of $f(x)$ is $\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right], k \in \mathbf{Z}$.