12. (14 points) In an ellipse, the chord intercepted by the ellipse on a line passing through a focus and perpendicular to the major axis is called the "latus rectum" of the ellipse. As shown in Figure 2, given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with foci $F_{1}$ and $F_{2}$, its eccentricity is $\frac{1}{2}$, and the length of the latus rectum is 3.
(1) Find the equation of the ellipse.
(2) A line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$, and $I_{1}$ and $I_{2}$ are the incenter of $\triangle F_{1} B F_{2}$ and $\triangle F_{1} A F_{2}$, respectively. Extend $B F_{2}$ to intersect the ellipse at point $M$. Find the ratio $p$ of the area of quadrilateral $F_{1} I_{2} F_{2} I_{1}$ to the area of $\triangle A F_{2} B$.
(3) Does there exist a fixed point $C$ on the $x$-axis such that $\overrightarrow{C M} \cdot \overrightarrow{C B}$ is a constant? If it exists, find the coordinates of point $C$; if not, explain the reason.
12. (1) From $\frac{c}{a}=\frac{1}{2}$, we get $a=2c$.

Given that the length of the latus rectum is 3, substituting $x=c$ into the ellipse equation yields $y^{2}=\frac{b^{4}}{a^{2}} \Rightarrow \frac{2 b^{2}}{a}=3$.

Solving this, we get $a=2, b=\sqrt{3}$.
Thus, the equation of the ellipse is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let the radius of the incircle of $\triangle F_{1} B F_{2}$ be $r$. Then $S_{\triangle F_{1} B F_{2}}=\frac{1}{2}\left(\left|B F_{1}\right|+\left|B F_{2}\right|+\left|F_{1} F_{2}\right|\right) r=3 r$, $S_{\triangle F_{1} I_{1} F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right| r=r$.

Therefore, $\frac{S_{\triangle F_{1} I_{1} F_{2}}}{S_{\triangle F_{1} B F_{2}}}=\frac{1}{3}$.
Similarly, $\frac{S_{\triangle F_{1} I_{2} F_{2}}}{S_{\triangle F_{1} F_{2}}}=\frac{1}{3}$.
Thus, $p=\frac{S_{\text {quadrilateral } F, l_{2} F_{2} I_{1}}}{S_{\triangle A F_{2} B}}=\frac{1}{3}$.
(3) Assume there exists a fixed point $C(n, 0)$ on the $x$-axis such that $\overrightarrow{C M} \cdot \overrightarrow{C B}$ is a constant.
Let the line $l_{B M}: x=m y+1$.
Solving the system of equations $\left\{\begin{array}{l}x=m y+1, \\ 3 x^{2}+4 y^{2}-12=0,\end{array}\right.$ we get
$$
\left(3 m^{2}+4\right) y^{2}+6 m y-9=0 \text {. }
$$

Let points $B\left(x_{1}, y_{1}\right), M\left(x_{2}, y_{2}\right)$. Then,
$y_{1}+y_{2}=-\frac{6 m}{3 m^{2}+4}, y_{1} y_{2}=-\frac{9}{3 m^{2}+4}$.
Thus, $\left(x_{1}-n\right)\left(x_{2}-n\right)=\left(m y_{1}+1-n\right)\left(m y_{2}+1-n\right)$
$=m^{2} y_{1} y_{2}+m(1-n)\left(y_{1}+y_{2}\right)+(1-n)^{2}$
$=-\frac{9 m^{2}}{3 m^{2}+4}-\frac{6 m^{2}(1-n)}{3 m^{2}+4}+(1-n)^{2}$
$=\frac{3 m^{2} n^{2}-12 m^{2}+4 n^{2}-8 n+4}{3 m^{2}+4}$.
Hence, $\overrightarrow{C M} \cdot \overrightarrow{C B}=\left(x_{1}-n\right)\left(x_{2}-n\right)+y_{1} y_{2}$
$=\frac{3 m^{2} n^{2}-12 m^{2}+4 n^{2}-8 n-5}{3 m^{2}+4}$
$=n^{2}-4-\frac{8 n-11}{3 m^{2}+4}$.
Since $\overrightarrow{C M} \cdot \overrightarrow{C B}$ is a constant, when $n=\frac{11}{8}$,
$$
\overrightarrow{C M} \cdot \overrightarrow{C B}=\left(\frac{11}{8}\right)^{2}-4=-\frac{135}{64} \text {. }
$$

Therefore, there exists a fixed point $C\left(\frac{11}{8}, 0\right)$ on the $x$-axis such that $\overrightarrow{C M} \cdot \overrightarrow{C B}$ is a constant.