12. As shown in the figure, let $k>0$ 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 $y=x+1$, and 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 $k \cdot k_{1}$;
(2) Prove that for any real number $k$, the line $M N$ always passes through a fixed point.
(1) As shown in the figure, let $P(x, y)$ be any point on $l_{1}$, and its symmetric point $P^{\prime}(y-1, x+1)$ with respect to the line $y=x+1$ lies on $l$. Therefore, we have $x+1=k(y-1)+1 \Rightarrow y=\frac{1}{k} x+1$,
which is the equation of the line $l_{1}$. So $k_{1}=\frac{1}{k} \Rightarrow k \cdot k_{1}=1$.
(2) Solving the system of equations for the line $l$ and the ellipse $E$:
$$
\left\{\begin{array}{l}
y=k x+1, \\
x^{2}+4 y^{2}=4
\end{array} \Rightarrow\left(1+4 k^{2}\right) x^{2}+8 k x=0 .\right.
$$

Thus, $x_{M}=\frac{-8 k}{1+4 k^{2}} \Rightarrow y_{M}=\frac{-8 k^{2}}{1+4 k^{2}}+1=\frac{1-4 k^{2}}{1+4 k^{2}} \Rightarrow M\left(\frac{-8 k}{1+4 k^{2}}, \frac{1-4 k^{2}}{1+4 k^{2}}\right)$. Replacing $k$ with $\frac{1}{k}$, we get $N\left(\frac{-8 k}{k^{2}+4}, \frac{k^{2}-4}{k^{2}+4}\right) \Rightarrow k_{M N}=-\frac{k^{2}+1}{3 k}$.
The line $M N: y-\frac{k^{2}-4}{k^{2}+4}=-\frac{k^{2}+1}{3 k}\left(x+\frac{8 k}{k^{2}+4}\right) \Rightarrow y=-\frac{k^{2}+1}{3 k} x-\frac{5}{3}$. Therefore, the line $M N$ always passes through the fixed point $B\left(0,-\frac{5}{3}\right)$.