15. Given that the center of the ellipse is the origin $O$, the foci are on the $x$-axis, a line with a slope of $\frac{1}{2}$ passing through the right focus $F$ intersects the ellipse at points $A$ and $B$, and $\overrightarrow{O A}+\overrightarrow{O B}$ is collinear with $a=$ $(-3,1)$.
(1) Find the eccentricity of the ellipse;
(2) Let $M$ be a point on the ellipse, and
$$
\overrightarrow{O M}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R}),
$$

Prove: $\lambda^{2}+\mu^{2}$ is a constant.
15. (1) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0), F(c, 0)$.
Then the line $l_{A B}: y=\frac{1}{2}(x-c)$.
Substituting into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, simplifying gives
$\left(a^{2}+4 b^{2}\right) x^{2}-2 a^{2} c x+a^{2} c^{2}-4 a^{2} b^{2}=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $x_{1}+x_{2}=\frac{2 a^{2} c}{a^{2}+4 b^{2}}, x_{1} x_{2}=\frac{a^{2} c^{2}-4 a^{2} b^{2}}{a^{2}+4 b^{2}}$.
Given $\overrightarrow{O A}+\overrightarrow{O B}=\left(x_{1}+x_{2}, y_{1}+y_{2}\right)$ is collinear with $\boldsymbol{a}=$ $(-3,1)$, we have
$3\left(y_{1}+y_{2}\right)+\left(x_{1}+x_{2}\right)=0$.
Also, $y_{1}=\frac{1}{2}\left(x_{1}-c\right), y_{2}=\frac{1}{2}\left(x_{2}-c\right)$, then
$3\left(\frac{1}{2}\left(x_{1}+x_{2}\right)-c\right)+\left(x_{1}+x_{2}\right)=0$
$\Rightarrow x_{1}+x_{2}=\frac{6}{5} c$
$\Rightarrow \frac{2 a^{2} c}{a^{2}+4 b^{2}}=\frac{6}{5} c$
$\Rightarrow a^{2}=6 b^{2}$
$\Rightarrow c=\sqrt{a^{2}-b^{2}}=\frac{\sqrt{30}}{6} a$
$\Rightarrow e=\frac{c}{a}=\frac{\sqrt{30}}{6}$.
(2) From (1), we know $a^{2}=6 b^{2}$.
Then the equation of the ellipse can be written as $x^{2}+6 y^{2}=6 b^{2}$.
Let $M(x, y)$.
From the given information,
$$
\begin{array}{l}
(x, y)=\lambda\left(x_{1}, y_{1}\right)+\mu\left(x_{2}, y_{2}\right) \\
\Rightarrow x=\lambda x_{1}+\mu x_{2}, y=\lambda y_{1}+\mu y_{2} .
\end{array}
$$

Since point $M(x, y)$ is on the ellipse, we have $\left(\lambda x_{1}+\mu x_{2}\right)^{2}+6\left(\lambda y_{1}+\mu y_{2}\right)^{2}=6 b^{2}$,
which simplifies to $\lambda^{2}\left(x_{1}^{2}+6 y_{1}^{2}\right)+\mu^{2}\left(x_{2}^{2}+6 y_{2}^{2}\right)+$
$2 \lambda \mu\left(x_{1} x_{2}+6 y_{1} y_{2}\right)=6 b^{2}$.
From (1), we know $a^{2}=\frac{6}{5} c^{2}, b^{2}=\frac{1}{5} c^{2}$, so $x_{1}+x_{2}=\frac{6}{5} c, x_{1} x_{2}=\frac{a^{2} c^{2}-4 a^{2} b^{2}}{a^{2}+4 b^{2}}=\frac{3}{25} c^{2}$.
Then $x_{1} x_{2}+6 y_{1} y_{2}$
$$
\begin{array}{l}
=x_{1} x_{2}+6 \times \frac{1}{4}\left(x_{1}-c\right)\left(x_{2}-c\right) \\
=\frac{5}{2} x_{1} x_{2}-\frac{3}{2}\left(x_{1}+x_{2}\right) c+\frac{3}{2} c^{2} \\
=\frac{3}{10} c^{2}-\frac{9}{5} c^{2}+\frac{3}{2} c^{2}=0 .
\end{array}
$$

Also, $x_{1}^{2}+6 y_{1}^{2}=6 b^{2}, x_{2}^{2}+6 y_{2}^{2}=6 b^{2}$, substituting into the equation
(1) gives $\lambda^{2}+\mu^{2}=1$.
Thus, $\lambda^{2}+\mu^{2}$ is a constant value of 1.