7. Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $E$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

The line passing through $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$, with $\left|A F_{1}\right|=3\left|B F_{1}\right|$, and $\cos \angle A F_{2} B=\frac{3}{5}$. The line $y=\frac{1}{2} x$ intersects the ellipse $E$ at points $P$ and $Q$. Points $C$ and $D$ are on the ellipse $E$ and are different from $P$ and $Q$, and the lines $P C$ and $Q D$ intersect at point $M$, while the lines $P D$ and $Q C$ intersect at point $N$. Then the slope of the line $M N$ is . $\qquad$
7. -1 .

Let $\left|F_{1} B\right|=k$. Then $k>0$, and
$$
\left|A F_{1}\right|=3 k,|A B|=4 k \text {. }
$$

By the definition of an ellipse, we have
$$
\left|A F_{2}\right|=2 a-3 k,\left|B F_{2}\right|=2 a-k \text {. }
$$

In $\triangle A B F_{2}$, by the cosine rule, we get
$$
\begin{array}{c}
|A B|^{2}=\left|A F_{2}\right|^{2}+\left|B F_{2}\right|^{2}- \\
2\left|A F_{2}\right|\left|B F_{2}\right| \cos \angle A F_{2} B \\
\Rightarrow(4 k)^{2}=(2 a-3 k)^{2}+(2 a-k)^{2}- \\
\frac{6}{5}(2 a-3 k)(2 a-k) \\
\Rightarrow(a+k)(a-3 k)=0 .
\end{array}
$$

Since $a+k>0$, we have $a=3 k$.
Then $\left|A F_{2}\right|=3 k=\left|A F_{1}\right|,\left|B F_{2}\right|=5 k$
$$
\begin{array}{l}
\Rightarrow\left|B F_{2}\right|^{2}=\left|F_{2} A\right|^{2}+|A B|^{2} \\
\Rightarrow F_{1} A \perp F_{2} A .
\end{array}
$$

Therefore, $\triangle A F_{1} F_{2}$ is an isosceles right triangle.
Thus, $a=\sqrt{2} b$.
Let $P\left(x_{0}, y_{0}\right), M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.
Then $Q\left(-x_{0},-y_{0}\right)$.
It is easy to see that $k_{C P} k_{C Q}=-\frac{b^{2}}{a^{2}}=-\frac{1}{2}$, that is,
$$
k_{M P} k_{N Q}=-\frac{b^{2}}{a^{2}}=-\frac{1}{2} \text {. }
$$

Similarly, $k_{N P} k_{M Q}=-\frac{1}{2}$.
Then $\left\{\begin{array}{l}\frac{y_{1}-y_{0}}{x_{1}-x_{0}} \cdot \frac{y_{2}+y_{0}}{x_{2}+x_{0}}=-\frac{1}{2}, \\ \frac{y_{1}+y_{0}}{x_{1}+x_{0}} \cdot \frac{y_{2}-y_{0}}{x_{2}-x_{0}}=-\frac{1}{2}\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array}{l}
y_{1} y_{2}-y_{0}\left(y_{2}-y_{1}\right)-y_{0}^{2} \\
=-\frac{1}{2}\left[x_{1} x_{2}-x_{0}\left(x_{2}-x_{1}\right)-x_{0}^{2}\right] \\
y_{1} y_{2}+y_{0}\left(y_{2}-y_{1}\right)-y_{0}^{2} \\
=-\frac{1}{2}\left[x_{1} x_{2}+x_{0}\left(x_{2}-x_{1}\right)-x_{0}^{2}\right]
\end{array}\right. \\
\Rightarrow 2 y_{0}\left(y_{2}-y_{1}\right)=-\frac{1}{2} \times 2 x_{0}\left(x_{2}-x_{1}\right) \\
\Rightarrow k_{M N}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\frac{1}{2} \cdot \frac{x_{0}}{y_{0}}=-1
\end{array}
$$