15. Given the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, its minor axis length is 4, and the eccentricity is $e_{1}$. 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 $e_{1} \cdot e_{2}=1$.
(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 $M, N$. Let the slopes of the lines $F M$ and $F N$ be $k_{1}, k_{2}$, 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.
(1) $m=n \Rightarrow e_{2}=\sqrt{2} \Rightarrow e_{1}=\frac{\sqrt{2}}{2} \Rightarrow b=c=2 \Rightarrow a=2 \sqrt{2}$.

So the equation of the ellipse $E$ is $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$.
(2) According to the problem, $F(2,0)$, the line $l: x=n y+4, M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, solving the system of equations
$$
\left\{\begin{array}{l}
x=n y+4, \\
x^{2}+2 y^{2}=8
\end{array} \Rightarrow\left(n^{2}+2\right) y^{2}+8 n y+8=0 \Rightarrow\left\{\begin{array}{l}
y_{1}+y_{2}=\frac{-8 n}{n^{2}+2}, \\
y_{1} y_{2}=\frac{8}{n^{2}+2} .
\end{array}\right.\right.
$$

Thus, $k_{1}+k_{2}=\frac{y_{1}}{x_{1}-2}+\frac{y_{2}}{x_{2}-2}=\frac{y_{1}}{n y_{1}+2}+\frac{y_{2}}{n y_{2}+2}=\frac{2 n y_{1} y_{2}+2\left(y_{1}+y_{2}\right)}{\left(n y_{1}+2\right)\left(n y_{2}+2\right)}$ $=\frac{1}{n^{2}+2} \cdot \frac{16 n-16 n}{\left(n y_{1}+2\right)\left(n y_{2}+2\right)}=0$. Therefore, $k_{1}+k_{2}$ is a constant 0.