(12) Given that the vertex of the parabola $C$ is at the origin, and the focus coordinates are $F(2,0)$, the coordinates of point $P$ are $(m, 0)(m \neq 0)$.
- (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 $N$. Does there exist 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 it exists, find the coordinates of point $T$; if not, explain the reason.
(12) (1) From the given conditions, the equation of the parabola $C$ is $y^{2}=8 x$, the equation of the line $l_{1}$ is $y=x-m$, and the coordinates of point $Q$ are $(-m, 0)$. 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 (1), we get $m>-2$. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=2(m+4)$, $x_{1} x_{2}=m^{2}$,
$$
\begin{aligned}
|A B| & =\sqrt{1+1^{2}} \cdot\left|x_{1}-x_{2}\right| \\
& =\sqrt{2} \cdot \sqrt{4(m+4)^{2}-4 m^{2}} \\
& =8 \sqrt{m+2} .
\end{aligned}
$$

Since the distance from point $Q(-m, 0)$ to the line $l_{1}$ is $d=\sqrt{2} \cdot|m|$, the area of $\triangle Q A B$ is
$$
S=\frac{1}{2} \cdot \sqrt{2} \cdot|m| \cdot 8 \sqrt{m+2}=4 \sqrt{2} \cdot \sqrt{m^{3}+2 m^{2}},
$$

where $-20$; when $-\frac{4}{3}<m<0$, $f^{\prime}(m)<0$.

Therefore, $f(m)$ is an increasing function on the interval $\left(-2,-\frac{4}{3}\right]$, and a decreasing function on the interval $\left[-\frac{4}{3}, 0\right)$. So when $m=-\frac{4}{3}$, $f(m)$ reaches its maximum value $\frac{32}{27}$. 
Thus, the maximum area of $\triangle Q A B$ is $\frac{32 \sqrt{3}}{9}$.
(2) The equation of $l_{2}$ is $y=k(x-m)$. From $\left\{\begin{array}{l}y=k(x-m), \\ y^{2}=8 x,\end{array}\right.$ we get
$$
k^{2} x^{2}-2\left(m k^{2}+4\right) x+k^{2} m^{2}=0 \text {. }
$$

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 $T$ exists, with coordinates $(t, 0)$. Since the acute angles formed by $T M$ and $T N$ with the $x$-axis are equal, we have $k_{T M}+k_{T N}=0$, i.e., $\frac{y_{3}}{x_{3}-t}+\frac{y_{4}}{x_{4}-t}=0$, i.e.,
$$
\begin{array}{c}
\frac{k\left(x_{3}-m\right)}{x_{3}-t}+\frac{k\left(x_{4}-m\right)}{x_{4}-t}=0, \\
2 x_{3} x_{4}-(m+t)\left(x_{3}+x_{4}\right)+2 m t=0 .
\end{array}
$$

Therefore,
$$
2 m^{2}-(m+t) \cdot \frac{2 m k^{2}+8}{k^{2}}+2 m t=0 \Rightarrow t=-m .
$$

Thus, the point $T$ that satisfies the conditions exists, with coordinates $(-m, 0)$.