25. In the ellipse $C: \frac{\left(x-x_{0}\right)^{2}}{a^{2}}+\frac{\left(y-y_{0}\right)^{2}}{b^{2}}=1(a>b$ $>0)$, $F_{1}$ is the left focus, $O^{\prime}$ is the center, $A, \mathscr{B}$ are its right vertex and upper vertex, $P$ is a point on $\bigcirc$, $P F_{\text {、 }}$ is perpendicular to the x-axis, and $O^{\prime} P / / / A B$.
(2) If the ellipse $C$ always passes through the point $Q(1,0)$, and one of the directrix equations is $x+2=0$, find the range of the semi-major axis $a$.

保留了原文的换行和格式。
25. 11) First, flatten $C$, making $\bigcirc^{\prime}$ coincide with the center. The equation of $C$ is
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text {. }
$$

Let $P\left(x^{\prime}, y^{\prime}\right)$,
where $x^{\prime}=-c$.
$$
\begin{array}{l}
\because P O / / B . A, \\
\therefore \frac{y^{\prime}}{x^{\prime}}=-\frac{b}{a}, y^{\prime}=\frac{b c}{a}, \\
X \because P \in C, \\
\therefore \frac{c^{2}}{a^{2}}+\frac{\left(\frac{b c}{a}\right)^{2}}{b^{2}}=1, \frac{c^{2}}{a^{2}}=\frac{1}{2}, \\
\therefore e=\frac{c}{a}=\frac{\sqrt{2}}{2} .
\end{array}
$$
(2) Restore $C$ to its original position, its equation is
$$
\frac{\left(x-x_{0}\right)^{2}}{a^{2}}+\frac{\left(y-y_{0}\right)^{2}}{b^{2}}=1 \text {. }
$$

From (1), we know $a^{2}=2 c^{2}$, and $b=c$,
$\therefore$ The equation of $C$ can be simplified to
$$
\begin{array}{l}
\frac{\left(x-x_{0}\right)^{2}}{2 c^{2}}+\frac{\left(v-y_{0}\right)^{2}}{c^{2}}=1 . \\
\because Q(1,0) \in C, \\
\therefore\left(1-x_{0}\right)^{2}-2 y_{0}^{2}=2 c^{2} .
\end{array}
$$

Also, since the directrix is $x=-2$ (left directrix), $x_{0}-\frac{a^{2}}{c}=-2, x_{0}=\frac{a^{2}}{c}-2$, substituting into (1) gives $\left[1-\left(\frac{a^{2}}{c}-2\right)\right]^{2}+2 y_{0}^{2}=2 c^{2}$,

which is $(3-2 c)^{2}+2 y_{0}^{2}=2 c^{2}$.
$$
\therefore 2 y_{0}^{2}=2 c^{2}-(3-2 c)^{2}=-2 c^{2}+12 c-9 \geqslant 0 \text {. }
$$

Solving, we get $3-\frac{3 \sqrt{2}}{2} \leqslant c \leqslant 3+\frac{3 \sqrt{2}}{2}$.
$$
\therefore 3 \sqrt{2}-3 \leqslant a \leqslant 3 \sqrt{2}+3 \text {. }
$$