2. (20 points) Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have an eccentricity of $\frac{1}{2}$, and the upper and lower endpoints of the ellipse's minor axis be $A$ and $B$, respectively. A circle centered at $A$ with the semi-major axis length of the ellipse as its radius intersects the ellipse at points $C$ and $D$. The y-coordinate of the midpoint of $C D$ is $6-3 \sqrt{3}$.
(1) Find the equation of the ellipse.
(2) A line $l$ passes through the right focus $F$ of the ellipse and is not perpendicular to the $x$-axis. $l$ intersects the ellipse at points $M$ and $N$. Let the point symmetric to $N$ with respect to the $x$-axis be $N^{\prime}$. Is the line $M N^{\prime}$ passing through a fixed point? If so, find the point; otherwise, explain the reason.
2. (1) From the problem, we have
$$
\left\{\begin{array}{l}
a^{2}=b^{2}+c^{2}, \\
e=\frac{c}{a}=\frac{1}{2}
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2c, \\
b=\sqrt{3}c.
\end{array}\right.\right.
$$

Thus, point $A(0, \sqrt{3}c)$, and the semi-major axis of the ellipse is $2c$.
Therefore, the equation of the circle with center $A$ and radius equal to the semi-major axis of the ellipse is
$$
x^{2}+(y-\sqrt{3}c)^{2}=(2c)^{2}.
$$

Combining the equation of the ellipse, and eliminating $x$, we get
$$
y^{2}+6\sqrt{3}cy-9c^{2}=0.
$$

The y-coordinate of the midpoint of $CD$ is $6-3\sqrt{3}$. By the symmetry of the ellipse and the circle, the y-coordinates of points $C$ and $D$ are also $6-3\sqrt{3}$.
Substituting $y=6-3\sqrt{3}$ into equation (1), we solve to get $c=1$.
Thus, $a=2$, $b=\sqrt{3}$, and the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $M(x_{1}, y_{1})$, $N(x_{2}, y_{2})$. Then $N^{\prime}(x_{2}, -y_{2})$.
From (1), we know the right focus of the ellipse is $F(1,0)$.
Note that the line $l$ is not perpendicular to the $x$-axis and passes through point $F$, so let the line $l: y=k(x-1)$.

When $k=0$, the line $l: y=0$, and points $N$ and $N^{\prime}$ coincide, so the line $MN^{\prime}$ coincides with $l$. This implies that if the line $MN^{\prime}$ passes through a fixed point, that point must lie on the $x$-axis.

Assume $k \neq 0$. Combining the equations of the line and the ellipse, and eliminating $y$, we get
$\left(4k^{2}+3\right)x^{2}-8k^{2}x+4k^{2}-12=0$.
Thus, $x_{1}+x_{2}=\frac{8k^{2}}{4k^{2}+3}$, and $x_{1}x_{2}$ in the equation of the line $MN^{\prime}$, $y-y_{1}=\frac{y_{1}+y_{2}}{x_{1}-x_{2}}(x-x_{1})$,

let $y=0$, then
$$
x=\frac{-y_{1}(x_{1}-x_{2})}{y_{1}+y_{2}}+x_{1}=\frac{x_{1}y_{2}+x_{2}y_{1}}{y_{1}+y_{2}}.
$$

Since points $M$ and $N$ are on the line $l$, we have
$$
y_{1}=k(x_{1}-1), y_{2}=k(x_{2}-1).
$$

Substituting into equation (3), we get
$$
x=\frac{2x_{1}x_{2}-(x_{1}+x_{2})}{x_{1}+x_{2}-2}.
$$

Substituting equation (2) into equation (4) and simplifying, we get $x=4$. Therefore, the line $MN^{\prime}$ always passes through the fixed point $(4,0)$.