13. (25 points) Given the function
$$
f(x)=4 \cos x \cdot \sin \left(x+\frac{7 \pi}{6}\right)+a
$$

has a maximum value of 2. Find:
(1) the value of $a$ and the smallest positive period of $f(x)$;
(2) the intervals where $f(x)$ is monotonically decreasing.
$$
\begin{array}{l}
\text { Three, 13. (1) } f(x)=4 \cos x \cdot \sin \left(x+\frac{7 \pi}{6}\right)+a \\
=2\left(\sin \left(2 x+\frac{7 \pi}{6}\right)+\sin \frac{7 \pi}{6}\right)+a \\
=-2 \sin \left(2 x+\frac{\pi}{6}\right)-1+a
\end{array}
$$

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, then
$$
1+a=2 \Rightarrow a=1 \text {. }
$$

Thus, 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 $2 x+\frac{\pi}{6} \in\left[-\frac{\pi}{2}+2 k \pi, \frac{\pi}{2}+2 k \pi\right](k \in \mathbf{Z})$.

Then $x \in\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z})$.
Therefore, the decreasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) .
$$