Five. (20 points) Given the semi-ellipse $\frac{x^{2}}{4}+y^{2}=1(y>0)$, two perpendicular lines are drawn through a fixed point $C(1,0)$ intersecting the ellipse at points $P$ and $Q$, respectively. Here, $O$ is the origin, and $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse.
(1) Find the minimum value of $\left|P F_{1}+P F_{2}\right|$;
(2) Does there exist a line such that the sum of the x-coordinates of $P F_{1}$ and $P F_{2}$ is minimized? If not, explain why.
(1) Since $P O$ is the median of $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}+P F_{2}=2 P O, \left|P F_{1}+P F_{2}\right|=2|P O| \text {. }
$$

Therefore, when $P$ is at the vertex of the minor axis, $\left|P F_{1}+P F_{2}\right|$ achieves its minimum value of 2.
(2) From the problem analysis, the slopes of the lines $C P$ and $C Q$ both exist. Let the slopes of the lines $C P$ and $C Q$ be
$$
k_{C P}=k, k_{C Q}=-\frac{1}{k} \text {. }
$$

Then the equations of these two lines are
$$
l_{C P}: y=k(x-1), l_{C Q}: y=-\frac{1}{k}(x-1) \text {. }
$$

Since $P F_{1}+P F_{2}=2 P O, Q F_{1}+Q F_{2}=2 Q O$, it is sufficient to satisfy $O P \perp O Q$.
Then $k_{O P} k_{O Q}=\frac{k\left(x_{P}-1\right)}{x_{P}} \cdot \frac{-\frac{1}{k}\left(x_{Q}-1\right)}{x_{Q}}=-1$
$$
\Rightarrow x_{P}+x_{Q}=1 \text {. }
$$

From the ellipse equation, we get $y_{P}=\frac{\sqrt{4-x_{P}^{2}}}{2}, y_{Q}=\frac{\sqrt{4-x_{Q}^{2}}}{2}$, then
$$
\begin{array}{l}
k_{O P} k_{Q Q}=\frac{\sqrt{4-x_{P}^{2}}}{2 x_{P}} \cdot \frac{\sqrt{4-x_{Q}^{2}}}{2 x_{Q}}=-1 \\
\Rightarrow \sqrt{16-4\left(x_{P}^{2}+x_{Q}^{2}\right)+x_{P}^{2} x_{Q}^{2}}=-4 x_{P} x_{Q} . \\
\text { Let } x_{P} x_{Q}=t \leqslant 0, \text { and } x_{P}+x_{Q}=1 .
\end{array}
$$

Let $x_{P} x_{Q}=t \leqslant 0$, and $x_{P}+x_{Q}=1$.
Thus $\sqrt{16-4(1-2 t)+t^{2}}=-4 t$
$$
\Rightarrow 15 t^{2}-8 t-12=0 \text {. }
$$

Solving, we get $t=-\frac{2}{3}$ or $t=\frac{6}{5}$ (discard).
Then $\left\{\begin{array}{l}x_{p}+x_{Q}=1, \\ x_{p} x_{Q}=-\frac{2}{3}\end{array}\right.$.
Solving, we get $\left\{\begin{array}{l}x_{P}=\frac{3+\sqrt{33}}{6}, \\ x_{Q}=\frac{3-\sqrt{33}}{6} ;\end{array}\left\{\begin{array}{l}x_{P}=\frac{3-\sqrt{33}}{6}, \\ x_{Q}=\frac{3+\sqrt{33}}{6}\end{array}\right.\right.$.
Therefore, there exist lines such that $\boldsymbol{P F} F_{1}+P F_{2}$ and $\boldsymbol{Q} F_{1}+Q F_{2}$ are perpendicular to each other.