12. Given a parabola $C$ with its vertex at the origin and focus at $F(2,0)$, and a point $P(m, 0)(m \neq 0)$.
(1) Let the line $l_{1}$ passing through point $P$ with a slope of 1 intersect the parabola $C$ at points $A$ and $B$. If $m<0$, and the point $P$ is symmetric to point $Q$ with respect to the origin, find the maximum value of the area of $\triangle Q A B$.
(2) Let the line $l_{2}$ passing through point $P$ with a slope of $k(k \neq 0)$ intersect the parabola $C$ at points $M$ and $N$. Is there a point $T$ on the $x$-axis such that the acute angles formed by $T M$ and $T N$ with the $x$-axis are equal? If such a point exists, find the coordinates of point $T$; if not, explain why.
12. (1) From the given conditions, the parabola $C: y^{2}=8 x$, the line $l_{1}: y=x-m$, and the point $Q(-m, 0)$.
From $\left\{\begin{array}{l}y=x-m, \\ y^{2}=8 x,\end{array}\right.$ we get
$x^{2}-2(m+4) x+m^{2}=0$.
From the discriminant $\Delta>0$ of equation (1), we get $m>-2$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then
$x_{1}+x_{2}=2(m+4), x_{1} x_{2}=m^{2}$.
Thus, $|A B|=\sqrt{2}\left|x_{1}-x_{2}\right|$
$=\sqrt{2} \times \sqrt{4\left(m+4\right)^{2}-4 m^{2}}=8 \sqrt{m+2}$.
Since the distance from point $Q(-m, 0)$ to the line $l_{1}$ is $d=\sqrt{2}|m|$, the area of $\triangle Q A B$ is
$S=\frac{1}{2} \times \sqrt{2}|m| \cdot 8 \sqrt{m+2}$
$=4 \sqrt{2} \sqrt{m^{3}+2 m^{2}}$ for $-2 < m < 0$;
Let $f(m)=4 \sqrt{2} \sqrt{m^{3}+2 m^{2}}$.
Then $f^{\prime}(m)=\frac{4 \sqrt{2}\left(3 m^{2}+4 m\right)}{2 \sqrt{m^{3}+2 m^{2}}}$.
When $-2 < m < -\frac{4}{3}$, $f^{\prime}(m)>0$;
When $-\frac{4}{3}<m<0$, $f^{\prime}(m)<0$.
Thus, $f(m)$ is an increasing function on the interval $\left(-2,-\frac{4}{3}\right]$ and a decreasing function on the interval $\left[-\frac{4}{3}, 0\right)$.
Therefore, when $m=-\frac{4}{3}$, $f(m)$ reaches its maximum value $\frac{32}{27}$.
Hence, the maximum area of $\triangle Q A B$ is $\frac{32 \sqrt{3}}{9}$.
(2) Let the line $l_{2}: y=k(x-m)$.
From $\left\{\begin{array}{l}y=k(x-m), \\ y^{2}=8 x,\end{array}\right.$, we get
$k^{2} x^{2}-2\left(m k^{2}+4\right) x+k^{2} m^{2}=0$.
Let $M\left(x_{3}, y_{3}\right)$ and $N\left(x_{4}, y_{4}\right)$. Then
$x_{3}+x_{4}=\frac{2 m k^{2}+8}{k^{2}}, x_{3} x_{4}=m^{2}$.
Assume point $T(t, 0)$ exists. Since the acute angles formed by $T M$ and $T N$ with the $x$-axis are equal, we have
$$
\begin{array}{l}
k_{T M}+k_{T N}=0 \\
\Rightarrow \frac{y_{3}}{x_{3}-t}+\frac{y_{4}}{x_{4}-t}=0 \\
\Rightarrow \frac{k\left(x_{3}-m\right)}{x_{3}-t}+\frac{k\left(x_{4}-m\right)}{x_{4}-t}=0 \\
\Rightarrow 2 x_{3} x_{4}-(m+t)\left(x_{3}+x_{4}\right)+2 m t=0 \\
\Rightarrow 2 m^{2}-(m+t) \frac{2 m k^{2}+8}{k^{2}}+2 m t=0
\end{array}
$$
$\Rightarrow t=-m$.
Therefore, the point $T$ that satisfies the condition exists, and its coordinates are $(-m, 0)$.