(9) Given, in $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, the circumradius of $\triangle A B C$ is $R=\sqrt{3}$, and it satisfies $\tan B+\tan C=\frac{2 \sin A}{\cos C}$.
(1) Find the size of angle $B$ and side $b$;
(2) Find the maximum area of $\triangle A B C$.
9) (1) From the given, we have $\frac{\sin B}{\cos B}+\frac{\sin C}{\cos C}=\frac{2 \sin A}{\cos C}$, which means
$$
\sin B \cos C+\sin C \cos B=2 \sin A \cos B,
$$

Therefore, $\sin (B+C)=2 \sin A \cos B$.
Since $A+B+C=180^{\circ}$, we have $\sin (B+C)=\sin A$.
Since $\sin A \neq 0$, we have $\cos B=\frac{1}{2}$. Since $0<B<\pi$, we have $B=\frac{\pi}{3}$. Given $R=\sqrt{3}$, we have $b=2 R \sin B=3$.
(2) By the cosine rule, we know $b^{2}=a^{2}+c^{2}-2 a c \cos B$, so
$$
9=a^{2}+c^{2}-2 a c \cos \frac{\pi}{3}=a^{2}+c^{2}-a c \geqslant 2 a c-a c .
$$

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

Thus, when $a=c=3$, $\left(S_{\triangle A B C}\right)_{\text {max }}=\frac{9}{4} \sqrt{3}$.