12. (20 points) Given the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$ with its right focus at $F$, the line $y=k(x-2)(k \neq 0)$ passing through $F$ intersects the ellipse at points $P$ and $Q$. If the midpoint of $PQ$ is $N$, and $O$ is the origin, the line $ON$ intersects the line $x=3$ at point $M$, find
(1) the size of $\angle M F Q$;
(2) the maximum value of $\frac{P Q}{M F}$.
12. (1) By combining the equations of the ellipse and the line, we get
$$
\left(3 k^{2}+1\right) x^{2}-12 k^{2} x+12 k^{2}-6=0 \text {. }
$$

Let $P\left(x_{p}, y_{p}\right), Q\left(x_{q}, y_{q}\right)$. Then
$$
x_{p}+x_{q}=\frac{12 k^{2}}{3 k^{2}+1}, x_{p} x_{q}=\frac{12 k^{2}-6}{3 k^{2}+1} \text {. }
$$

Thus, $y_{p}+y_{q}=k\left(x_{p}+x_{q}\right)-4 k=\frac{-4 k}{3 k^{2}+1}$.
The midpoint of $P Q$ is $N\left(\frac{6 k^{2}}{3 k^{2}+1}, \frac{-2 k}{3 k^{2}+1}\right)$, so $k_{O N}=-\frac{1}{3 k}$.
The line $O N$ intersects the line $x=3$ at point $M$, thus $M\left(3,-\frac{1}{k}\right)$.
Therefore, $k_{M F}=-\frac{1}{k}$.
Hence, $k_{M F} k_{P Q}=-1 \Rightarrow M F \perp P Q$
$$
\Rightarrow \angle M F Q=\frac{\pi}{2} \text {. }
$$
(2) Note that,
$$
\begin{array}{l}
I=\left(\frac{P Q}{M F}\right)^{2}=\frac{\left(x_{p}-x_{q}\right)^{2}+k^{2}\left(x_{p}-x_{q}\right)^{2}}{1+\frac{1}{k^{2}}} \\
=k^{2}\left(\left(x_{p}+x_{q}\right)^{2}-4 x_{p} x_{q}\right) \\
=24 k^{2} \frac{k^{2}+1}{\left(3 k^{2}+1\right)^{2}} \text {. } \\
\end{array}
$$

Let $u=3 k^{2}+1$. Then
$$
\begin{array}{l}
I=8 \frac{(u-1)(u+2)}{3 u^{2}} \\
=-\frac{16}{3}\left(\left(\frac{1}{u}-\frac{1}{4}\right)^{2}-\frac{9}{16}\right) \leqslant 3 .
\end{array}
$$

When $u=4$, i.e., $k= \pm 1$, $I$ reaches its maximum value. In summary, the maximum value of $\frac{P Q}{M F}$ is $\sqrt{3}$.