13. Given $\triangle A B C$ with internal angles $\angle A, \angle B, \angle C$ and their opposite sides $a, b, c$ respectively, and it is known that $1+\frac{\tan B}{\tan A}=\frac{2 c}{\sqrt{3} a}$.
(1) Find $\angle B$;
(2) If $\triangle A B C$ is an acute triangle, and $a=2$, find the range of the area of $\triangle A B C$.
13. (1) Since $1+\frac{\tan B}{\tan A}=\frac{2 c}{\sqrt{3} a}$, by the Law of Sines, we have
$$
\begin{array}{l}
1+\frac{\sin B \cdot \cos A}{\cos B \cdot \sin A}=\frac{\sin (A+B)}{\cos B \cdot \sin A} \\
=\frac{2 \sin C}{\sqrt{3} \sin A} .
\end{array}
$$

Since $\angle A+\angle B+\angle C=\pi$, then $\sin (A+B)=\sin C$.
Thus, $\frac{\sin C}{\cos B \cdot \sin A}=\frac{2 \sin C}{\sqrt{3} \sin A}$.
Given that $\sin A \neq 0, \sin C \neq 0$, we have $\cos B=\frac{\sqrt{3}}{2}$.
Since $\angle B \in(0, \pi)$, we have $\angle B=\frac{\pi}{6}$.
(2) From (1), we know $\angle B=\frac{\pi}{6}$.

Therefore, $\angle A+\angle C=\frac{5 \pi}{6}$.
Since $\triangle A B C$ is an acute triangle, we have
$$
\left\{\begin{array}{l}
0<\angle A<\frac{\pi}{2} \\
0<\angle C<\frac{\pi}{2}
\end{array} \Rightarrow \frac{\pi}{3}<\angle A<\frac{\pi}{2} \Rightarrow \frac{\sqrt{3}}{3}<\tan A<\sqrt{3}\right.
$$
Thus,
$$
\frac{\sqrt{3}}{2}<\frac{1}{2 \tan A}+\frac{\sqrt{3}}{2}<\frac{2 \sqrt{3}}{3}
$$
$$
\Rightarrow \frac{\sqrt{3}}{2}<S_{\triangle A B C}<\frac{2 \sqrt{3}}{3}.
$$
Therefore, the range of $S_{\triangle A B C}$ is $\left(\frac{\sqrt{3}}{2}, \frac{2 \sqrt{3}}{3}\right)$.