16. (12 points) Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

the chord $AB$ passes through the point $F_{2}$, and
$$
\left|A F_{2}\right|=2\left|F_{2} B\right|, \tan \angle A F_{1} B=\frac{3}{4} .
$$
(1) Find the eccentricity $e$ of the ellipse;
(2) If the area of $\triangle F_{1} F_{2} B$ is 2, find the equation of the ellipse.
16. (1) Let $\left|A F_{2}\right|=2\left|F_{2} B\right|=2 k$.

From $\left|A F_{1}\right|+\left|A F_{2}\right|=2 a$,
$\left|B F_{1}\right|+\left|B F_{2}\right|=2 a$,
we get $\left|A F_{1}\right|=2 a-2 k,\left|B F_{1}\right|=2 a-k$.
Since $\tan \angle A F_{1} B=\frac{3}{4}$, we have
$\cos \angle A F_{1} B=\frac{4}{5}$.
In $\triangle A F_{1} B$, by the cosine rule we get
$|A B|^{2}=\left|A F_{1}\right|^{2}+\left|B F_{1}\right|^{2}-2\left|A F_{1}\right|\left|B F_{1}\right| \cos \angle A F_{1} B$,

which is $(3 k)^{2}=(2 a-2 k)^{2}+(2 a-k)^{2}-$
$$
2(2 a-2 k)(2 a-k) \cdot \frac{4}{5} \text {. }
$$

Simplifying, we get $(a-3 k)(2 a+3 k)=0$.
Since $2 a+3 k>0$, we have $a=3 k$.
Thus, $\left|A F_{1}\right|=2 a-2 k=4 k,|A B|=3 k$, $\left|B F_{1}\right|=2 a-k=5 k$.
Therefore, $\left|A F_{1}\right|^{2}+|A B|^{2}=\left|B F_{1}\right|^{2}$.
Hence, $\angle F_{1} A F_{2}=90^{\circ}$.
In the right triangle $\triangle A F_{1} F_{2}$, let $\left|F_{1} F_{2}\right|=2 c$. Then $\left|A F_{1}\right|^{2}+\left|A F_{2}\right|^{2}=\left|F_{1} F_{2}\right|^{2}$,
which is $(4 k)^{2}+(2 k)^{2}=4 c^{2}$.
Solving, we get $c=\sqrt{5} k$.
Therefore, $e=\frac{c}{a}=\frac{\sqrt{5}}{3}$.
(2) From $\frac{S_{\triangle F_{1} F_{2}}}{S_{\triangle F_{1} F_{2} B}}=\frac{\left|A F_{2}\right|}{\left|F_{2} B\right|}=2$, we get
$S_{\triangle A F_{1} F_{2}}=\frac{1}{2}\left|A F_{1}\right|\left|A F_{2}\right|=2 S_{\triangle F_{1} F_{2} B}=4$,

which is $\frac{1}{2} \times 4 k \times 2 k=4 \Rightarrow k=1$.
Thus, $a=3, c=\sqrt{5}, b=\sqrt{a^{2}-c^{2}}=2$.
Therefore, the equation of the ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$.