12. Given that $F$ is the right focus of the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, and the ratio of the distance from any point $P$ on the ellipse $C$ to the point $F$ to the distance from point $P$ to the line $l: x=m$ is $\frac{1}{2}$.
(1) Find the equation of the line $l$.
(2) Let $A$ be the left vertex of the ellipse $C$. A line passing through the point $F$ intersects the ellipse $C$ at points $D$ and $E$. The lines $A D$ and $A E$ intersect the line $l$ at points $M$ and $N$, respectively. Does the circle with diameter $M N$ always pass through a fixed point? If so, find the coordinates of the fixed point; if not, explain the reason.
12. (1) From the problem, we know $F(1,0)$.

Let $P(x, y)$ be any point on the ellipse $C$.
Thus $\frac{\sqrt{(x-1)^{2}+y^{2}}}{|x-m|}=\frac{1}{2}$
$\Rightarrow 4(x-1)^{2}+4 y^{2}=(x-m)^{2}$.
Substituting $4 y^{2}=12-3 x^{2}$ into the above equation and simplifying, we get
$(8-2 m) x+m^{2}-16=0$.
Since $P(x, y)$ is any point on the ellipse, the equation (1) holds for $-2 \leqslant x \leqslant 2$.
Thus $8-2 m=0$, and $m^{2}-16=0$.
Solving, we get $m=4$.
Therefore, the equation of the line $l$ is $x=4$.
(2) It is easy to see that the slope of line $DE$ is not 0.
Let $l_{DE}: x=t y+1$.
Combining with the equation of the ellipse, we get
$$
\left(3 t^{2}+4\right) y^{2}+6 t y-9=0 \text {. }
$$

Let $D\left(x_{1}, y_{1}\right), E\left(x_{2}, y_{2}\right)$. Then
$y_{1}+y_{2}=\frac{-6 t}{3 t^{2}+4}, y_{1} y_{2}=\frac{-9}{3 t^{2}+4}$.
Given $A(-2,0)$, we have
$l_{AD}: y-0=\frac{y_{1}-0}{x_{1}+2}(x+2)$.
Thus, $M\left(4, \frac{6 y_{1}}{x_{1}+2}\right)$.
Similarly, $N\left(4, \frac{6 y_{2}}{x_{2}+2}\right)$.
By symmetry, if a fixed point exists, it lies on the x-axis. Let point $G(n, 0)$ be on the circle with diameter $MN$.
Then $\overrightarrow{GM} \cdot \overrightarrow{GN}$
$=\left(4-n, \frac{6 y_{1}}{x_{1}+2}\right) \cdot\left(4-n, \frac{6 y_{2}}{x_{2}+2}\right)$
$=(4-n)^{2}+\frac{36 y_{1} y_{2}}{\left(x_{1}+2\right)\left(x_{2}+2\right)}$
$=(4-n)^{2}+\frac{36 y_{1} y_{2}}{t^{2} y_{1} y_{2}+3 t\left(y_{1}+y_{2}\right)+9}$
$=(4-n)^{2}+\frac{36 \times(-9)}{-9 t^{2}+3 t(-6 t)+9\left(3 t^{2}+4\right)}$
$=0$
$\Rightarrow(4-n)^{2}-9=0$
$\Rightarrow n=1$ or 7.
Thus, the circle with diameter $MN$ always passes through two fixed points on the x-axis, $(1,0),(7,0)$.