16. (25 points) Let the hyperbola $\Gamma: \frac{x^{2}}{4}-\frac{y^{2}}{5}=1$ have its left and right foci as $F_{1}$ and $F_{2}$, respectively. Let $P$ be a moving point on the hyperbola and in the first quadrant, and let the centroid and incenter of $\triangle P F_{1} F_{2}$ be $G$ and $I$, respectively.
(1) Does there exist a point $P$ such that $IG \parallel F_{1} F_{2}$? If so, find the point $P$; if not, explain why.
(2) Let $A$ be the left vertex of the hyperbola $\Gamma$, and let the line $l$ pass through the right focus $F_{2}$ and intersect the hyperbola $\Gamma$ at points $M$ and $N$. If the slopes of $AM$ and $AN$ are $k_{1}$ and $k_{2}$, respectively, and $k_{1}+k_{2}=-\frac{1}{2}$, find the equation of the line $l$.
16. (1) Suppose there exists a point $P\left(x_{0}, y_{0}\right)\left(x_{0}>0, y_{0}>0\right)$ such that $IG // F_{1}F_{2}$.
Since $G$ is the centroid of $\triangle P F_{1} F_{2}$, we have $G\left(\frac{x_{0}}{3}, \frac{y_{0}}{3}\right)$.
And $I$ is the incenter of $\triangle P F_{1} F_{2}$, let the inradius of $\triangle P F_{1} F_{2}$ be $r$. Then
$$
\begin{array}{l}
S_{\triangle P F_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{0}\right| \\
=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+\left|F_{1} F_{2}\right|\right) r \\
\Rightarrow \frac{1}{2} \times 2 c\left|y_{0}\right|=\frac{1}{2}\left(\left|P F_{1}\right|+\left|P F_{2}\right|+2 c\right) r \\
\Rightarrow r=\frac{2 c y_{0}}{\left|P F_{1}\right|+\left|P F_{2}\right|+2 c} .
\end{array}
$$

Since $IG // F_{1}F_{2}$, we have
$$
\begin{array}{l}
\frac{2 c y_{0}}{\left|P F_{1}\right|+\left|P F_{2}\right|+2 c}=\frac{y_{0}}{3} \\
\Rightarrow\left|P F_{1}\right|+\left|P F_{2}\right|=4 c=12 .
\end{array}
$$

Also, $\left|P F_{1}\right|-\left|P F_{2}\right|=2 a=4$, solving this gives
$$
\left|P F_{2}\right|=4 \text {. }
$$

Thus, $\left\{\begin{array}{l}\left(x_{0}-3\right)^{2}+y_{0}^{2}=16, \\ \frac{x_{0}^{2}}{4}-\frac{y_{0}^{2}}{5}=1 .\end{array}\right.$
Since point $P$ is in the first quadrant, solving this gives $x_{0}=4, y_{0}=\sqrt{15}$ (discard the negative solution).

Therefore, there exists a point $P(4, \sqrt{15})$ such that $IG // F_{1}F_{2}$.
(2) By the problem, let the equation of the line passing through point $F_{2}(3,0)$ be $y=k(x-3)$, and the line intersects the ellipse at points
$$
\begin{array}{l}
M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right) . \\
\quad \text { By }\left\{\begin{array}{l}
y=k(x-3), \\
5 x^{2}-4 y^{2}=20
\end{array}\right. \\
\quad \Rightarrow\left(5-4 k^{2}\right) x^{2}+24 k^{2} x-36 k^{2}-20=0 .
\end{array}
$$

By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{24 k^{2}}{4 k^{2}-5}, x_{1} x_{2}=\frac{36 k^{2}+20}{4 k^{2}-5} . \\
\text { Also, } k_{1}+k_{2}=\frac{y_{1}}{x_{1}+2}+\frac{y_{2}}{x_{2}+2} \\
=k\left(\frac{x_{1}-3}{x_{1}+2}+\frac{x_{2}-3}{x_{2}+2}\right) \\
=k\left[2-5\left(\frac{1}{x_{1}+2}+\frac{1}{x_{2}+2}\right)\right], \\
\frac{1}{x_{1}+2}+\frac{1}{x_{2}+2}=\frac{x_{1}+x_{2}+4}{x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4} \\
=\frac{24 k^{2}+4\left(4 k^{2}-5\right)}{36 k^{2}+20+48 k^{2}+4\left(4 k^{2}-5\right)}=\frac{2 k^{2}-1}{5 k^{2}}, \\
\text { Then } k_{1}+k_{2}=k\left(2-5 \times \frac{2 k^{2}-1}{5 k^{2}}\right)=\frac{1}{k}=-\frac{1}{2} \\
\Rightarrow k=-2 .
\end{array}
$$

Therefore, the equation of the line $l$ is $y=-2 x+6$.