Four. (Full marks 20 points) Given point $A(1,1)$ and line $l: x=3$ in the coordinate plane, let the moving point $P$ have a distance $m$ to point $A$ and a distance $n$ to line $l$. If $m+n=4$,
(1) Find the equation of the trajectory of point $P$ and draw the graph;
(2) Draw a line $s$ through $A$ with an inclination angle $\alpha$, intersecting the trajectory $C$ at points $Q$ and $R$. Let $f(\alpha)=|QR|$. Find the expression for $f(\alpha)$ and the value of $\alpha$ when $f(\alpha)$ is maximized and minimized.
(1) Let $P(x, y)$, then
$$
m=\sqrt{(x-1)^{2}+(y-1)^{2}}, n=|x-3| \text {. }
$$

From the condition $m+n=4$ we have
$$
\sqrt{(x-1)^{2}+(y-1)^{2}}+|x-3|=4 \text {. }
$$

It is easy to see that $0 \leqslant x \leqslant 4$ (otherwise $m+n>4$ ).
When $0 \leqslant x \leqslant 3$, the equation is
$$
(x-1)^{2}+(y-1)^{2}=(x+1)^{2} \text {, }
$$

which simplifies to $(y-1)^{2}=4 x$.
When $3 \leqslant x \leqslant 4$, the equation is
$$
(x-1)^{2}+(y-1)^{2}=(7-x)^{2} \text {, }
$$

which simplifies to $(y-1)^{2}=-12(x-4)$.
The trajectory is the shape shown in Figure 4, consisting of two parabolas.
$A(1,1)$
is the common focus of the parabolas, and $B, B^{\prime}$
are the intersection points of the two parabolas. We find $B(3$,
$$
\begin{array}{l}
2 \sqrt{3}+1), \\
B^{\prime}(3,-2 \sqrt{3}+1), \\
\angle B A x^{\prime}=\frac{\pi}{3} .
\end{array}
$$
(2) Establish a polar coordinate system with the line $A x^{\prime}$, parallel to the $x$-axis and passing through $A$, as the polar axis, and $A$ as the pole. Then
the polar equation of parabola (1) is: $\rho=\frac{2}{1-\cos \theta}$,
the polar equation of parabola (2) is: $\rho=\frac{6}{1+\cos \theta}$.
Depending on the size of the inclination angle $\alpha$ of the line $s$, there are three cases:
If $0 \leqslant \alpha \leqslant \frac{\pi}{3}$, then
$$
\begin{aligned}
f(\alpha) & =\rho_{1}+\rho_{2}=\frac{6}{1+\cos \alpha}+\frac{2}{1-\cos (\pi+\alpha)} \\
& =\frac{8}{1+\cos \alpha}
\end{aligned}
$$

If $\frac{\pi}{3} \leqslant \alpha \leqslant \frac{2 \pi}{3}$, then
$$
\begin{aligned}
f(\alpha) & =\rho_{1}+\rho_{2}=\frac{2}{1-\cos \alpha}+\frac{2}{1-\cos (\pi+\alpha)} \\
& =\frac{4}{\sin ^{2} \alpha}
\end{aligned}
$$

If $\frac{2 \pi}{3} \leqslant \alpha<\pi$, then
$$
\begin{aligned}
f(\alpha) & =\rho_{1}+\rho_{2}=\frac{2}{1-\cos \alpha}+\frac{6}{1+\cos (\pi+\alpha)} \\
& =\frac{8}{1-\cos \alpha} .
\end{aligned}
$$

It is known that when $\alpha=0$ or $\frac{\pi}{2}$, $f(\alpha)$ has a minimum value of 4;
when $\alpha=\frac{\pi}{3}$ or $\frac{2 \pi}{3}$, $f(\alpha)$ has a maximum value of $\frac{16}{3}$.