9. (14 points) In $\triangle ABC$, it is known that the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are $a$, $b$, and $c$ respectively, the circumradius of $\triangle ABC$ is $R=\sqrt{3}$, and it satisfies
$\tan B + \tan C = \frac{2 \sin A}{\cos C}$.
Find: (1) $\angle B$, $b$;
(2) the maximum area of $\triangle ABC$.
(1) From the given,
$$
\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}=\frac{2 \sin A}{\cos C},
$$

which means $\sin B \cdot \cos C+\sin C \cdot \cos B=2 \sin A \cdot \cos B$.
Thus, $\sin (B+C)=2 \sin A \cdot \cos B$.
Since $\angle A+\angle B+\angle C=\pi$, we have
$\sin (B+C)=\sin A$.
Given $\sin A \neq 0$, then $\cos B=\frac{1}{2}$.
Since $0<\angle B<\pi$, it follows that $\angle B=\frac{\pi}{3}$.
Given $R=\sqrt{3}$, we have $b=2 R \sin B=3$.
(2) By the cosine rule,
$$
b^{2}=a^{2}+c^{2}-2 a c \cos B.
$$

Thus, $9=a^{2}+c^{2}-2 a c \cos \frac{\pi}{3}$
$$
=a^{2}+c^{2}-a c \geqslant 2 a c-a c.
$$

Therefore, $a c \leqslant 9$, with equality holding if and only if $a=c=3$.
Hence, $S_{\triangle A B C}=\frac{1}{2} a c \sin B$
$$
\leqslant \frac{1}{2} \times 9 \sin \frac{\pi}{3}=\frac{9 \sqrt{3}}{4},
$$

which means that when $a=c=3$, the maximum value of $S_{\triangle A B C}$ is $\frac{9 \sqrt{3}}{4}$.