12. (20 points) As shown in Figure 2, given $k>0$ and $k \neq 1$, the line $l: y=k x+1$ is symmetric to the line $l_{1}: y=k_{1} x+1$ with respect to the line $y=x+1$. The lines $l$ and $l_{1}$ intersect the ellipse $E$: $\frac{x^{2}}{4}+y^{2}=1$ at points $A, M$ and $A, N$ respectively.
(1) Find the value of $k k_{1}$;
(2) Prove that for any $k$, the line $M N$ always passes through a fixed point.
12. (1) It is easy to know that the intersection point of line $l$ and $l_{1}$ is $A(0,1)$. Let $P(x, y)$ be any point on line $l$ other than point $A(0,1)$, and $P_{0}\left(x_{0}, y_{0}\right)$ be the symmetric point of $P$ with respect to the line $y=x+1$.
$$
\begin{aligned}
\text { By } \frac{y+y_{0}}{2}=\frac{x+x_{0}}{2}+1 \\
\Rightarrow y-x=x_{0}-y_{0}+2 . \\
\text { By } \frac{y-y_{0}}{x-x_{0}}=-1 \\
\Rightarrow y+x=y_{0}+x_{0} .
\end{aligned}
$$

Solving equations (1) and (2) simultaneously, we get
$$
x=y_{0}-1, y=x_{0}+1 \text {. }
$$

Substituting into the equation of line $l$ gives
$$
x_{0}=k\left(y_{0}-1\right) \text {. }
$$

Since point $P_{0}\left(x_{0}, y_{0}\right)$ lies on line $l_{1}$, we have
$$
y_{0}=k_{1} x_{0}+1 \Rightarrow y_{0}-1=k_{1} x_{0} \text {. }
$$

Thus, $x_{0}=k k_{1} x_{0}$.
Therefore, by $x_{0} \neq 0$, we get $k k_{1}=1$.
(2) Let points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$.

From $\left\{\begin{array}{l}y_{1}=k x_{1}+1, \\ \frac{x_{1}^{2}}{4}+y_{1}^{2}=1\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow\left(4 k^{2}+1\right) x_{1}^{2}+8 k x_{1}=0 \\
\Rightarrow x_{1}=\frac{-8 k}{4 k^{2}+1}, y_{1}=\frac{1-4 k^{2}}{4 k^{2}+1} .
\end{array}
$$

Similarly, $x_{2}=\frac{-8 k_{1}}{4 k_{1}^{2}+1}, y_{2}=\frac{1-4 k_{1}^{2}}{4 k_{1}^{2}+1}$.
Also, by $k k_{1}=1$, we get
$$
x_{2}=\frac{-8 k}{4+k^{2}}, y_{2}=\frac{k^{2}-4}{4+k^{2}} \text {. }
$$

Then the slope of line $M N$ is
$$
\begin{array}{l}
k_{M N}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{\frac{1-4 k^{2}}{4 k^{2}+1}-\frac{k^{2}-4}{4+k^{2}}}{\frac{-8 k}{4 k^{2}+1}-\frac{-8 k}{4+k^{2}}} \\
=\frac{8-8 k^{4}}{8 k\left(3 k^{2}-3\right)}=-\frac{k^{2}+1}{3 k} .
\end{array}
$$

Thus, the equation of line $M N$ is
$y-\frac{1-4 k^{2}}{4 k^{2}+1}=-\frac{k^{2}+1}{3 k}\left(x-\frac{-8 k}{4 k^{2}+1}\right)$
$\Rightarrow y=-\frac{k^{2}+1}{3 k} x-\frac{5}{3}$.
Therefore, for any $k$, line $M N$ always passes through the fixed point
$$
\left(0,-\frac{5}{3}\right) \text {. }
$$