15. (14 points) Let the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, with the minor axis being 4, and the eccentricity being $e_{1}$. The hyperbola $\frac{x^{2}}{m}-\frac{y^{2}}{n}=1$ $(m>0, n>0)$ has asymptotes $y= \pm x$, and the eccentricity is $e_{2}$, and $e_{1} e_{2}=1$.
(1) Find the equation of the ellipse $E$.
(2) Let the right focus of the ellipse $E$ be $F$, and draw a line through the point $G(4,0)$ with a non-zero slope, intersecting the ellipse $E$ at points $M$ and $N$. Let the slopes of the lines $F M$ and $F N$ be $k_{1}$ and $k_{2}$, respectively. Determine whether $k_{1}+k_{2}$ is a constant value. If it is a constant value, find the value; if not, explain the reason.
15. As shown in Figure 5.
(1) From the problem, we know $2 b=4, b=2, \frac{n}{m}=1$.
Then the eccentricity of the hyperbola is
$$
e_{2}=\sqrt{1+\frac{n}{m}}=\sqrt{2} \text {. }
$$

Thus, the eccentricity of the ellipse is
$$
\begin{aligned}
e_{1} & =\frac{1}{e_{2}}=\frac{c}{a}=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{\sqrt{2}}{2} \\
\Rightarrow a & =2 \sqrt{2} .
\end{aligned}
$$

Therefore, the standard equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$.
(2) Let $l_{M N}: y=k(x-4)(k \neq 0)$.

Combining with the equation of the ellipse and eliminating $y$, we get
$$
\left(1+2 k^{2}\right) x^{2}-16 k^{2} x+32 k^{2}-8=0 \text {. }
$$

Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{16 k^{2}}{2 k^{2}+1}, x_{1} x_{2}=\frac{32 k^{2}-8}{2 k^{2}+1}, \\
k_{1}+k_{2}=\frac{y_{1}}{x_{1}-2}+\frac{y_{2}}{x_{2}-2} \\
=\frac{k\left(x_{1}-4\right)}{x_{1}-2}+\frac{k\left(x_{2}-4\right)}{x_{2}-2} \\
=k \cdot \frac{\left(x_{1}-4\right)\left(x_{2}-2\right)+\left(x_{2}-4\right)\left(x_{1}-2\right)}{\left(x_{1}-2\right)\left(x_{2}-2\right)} \\
=k \cdot \frac{2 x_{1} x_{2}-6\left(x_{1}+x_{2}\right)+16}{\left(x_{1}-2\right)\left(x_{2}-2\right)} .
\end{array}
$$

Substituting $x_{1}+x_{2}=\frac{16 k^{2}}{2 k^{2}+1}, x_{1} x_{2}=\frac{32 k^{2}-8}{2 k^{2}+1}$ into
the above equation, we get
$$
\begin{array}{l}
2 x_{1} x_{2}-6\left(x_{1}+x_{2}\right)+16=0 \\
\Rightarrow k_{1}+k_{2}=0 .
\end{array}
$$