1. From the given information, the parabola's equation is $y^2 = 4x$, and $P$ is $(-1, 0)$.
Let $A(x\_1, y\_1)$ and $B(x\_2, y\_2)$. The line $l$ has the equation $x = my - 1$. Substituting this into the parabola's equation, we get:
$$y^2 - 4my + 4 = 0 \qquad \Delta = 16m^2 - 16 > 0 \qquad m^2 > 1$$
$$\begin{cases} y_1 + y_2 = 4m \\ y_1y_2 = 4 \end{cases}$$
Suppose point $T(a, 0)$ meets the requirement. Then, the slopes $k_{AT}$ and $k_{BT}$ satisfy:
$$\frac{y_1}{x_1 - a} + \frac{y_2}{x_2 - a} = \frac{2my_1y_2 - (1+a)(y_1 - y_2)}{(x_1 - a)(x_2 - a)} = \frac{8m - 4m(1+a)}{(x_1 - a)(x_2 - a)} = 0$$
$$\therefore 8m - 4m(1+a) = 0 \qquad \therefore a = 1$$
Hence, point $T(1, 0)$ exists.

2. The area of triangle $AOB$ is given by:
$$S_{\triangle AOB} = \frac{1}{2}|\overrightarrow{OA}||\overrightarrow{OB}|\sin\theta = \frac{5}{2}$$
$$\therefore |\overrightarrow{OA}||\overrightarrow{OB}| = \frac{5}{\sin\theta}$$
$$\overrightarrow{OA} \cdot \overrightarrow{OB} = x_1x_2 + y_1y_2 = \frac{y_1^2}{4} \cdot \frac{y_2^2}{4} + y_1y_2 = \frac{4^2}{16} + 4 = 5$$
$$\therefore \cos\angle{AOB} = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}||\overrightarrow{OB}|} = \sin\angle{AOB}$$
$$\therefore \tan\angle{AOB} = 1$$
$$\therefore \angle{AOB} = \boxed{\frac{\pi}{4}}$$