10. Given the standard equation of ellipse $C$ as $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>1)$, point $P$ is a moving point on the unit circle, points $A, F$ are the right vertex and right focus of ellipse $C$ respectively, and when $\angle P O A=\frac{\pi}{3}$, $|P F|=1,|P A|=\sqrt{3}$.
(1) Find the standard equation of ellipse $C$;
(2) Draw a line $l$ through point $P$ intersecting ellipse $C$ at points $M, N$, and satisfying $P$ as the midpoint of segment $M N$. Then draw the tangents to ellipse $C$ at points $M, N$ intersecting at point $Q$. Find the range of $|O Q|$ as point $P$ moves on the unit circle.
(1) $\angle P O A=\frac{\pi}{3},|P F|=|O P|=1 \Rightarrow \triangle P O F$ is an equilateral triangle, then $|O F|=1$,

By the cosine rule, $|P A|^{2}=|O P|^{2}+|O A|^{2}-|O P| \cdot|O A| \Rightarrow|O A|^{2}-|O A|-2=0$
$\Rightarrow|O A|=2$, thus $a=2, c=1, b=\sqrt{3}$. Therefore, the equation of the ellipse $C$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
(2) Let $Q(s, t), M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then
$$
M Q: \frac{x_{1} x}{4}+\frac{y_{1} y}{3}=1, N Q: \frac{x_{2} x}{4}+\frac{y_{2} y}{3}=1 \text {, }
$$

Thus $M N: \frac{s x}{4}+\frac{t y}{3}=1 \Rightarrow 3 s x+4 t y=12$,
Solving the system of equations $\left\{\begin{array}{l}3 s x+4 t y=12, \\ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1\end{array}\right.$
$$
\Rightarrow\left(3 s^{2}+4 t^{2}\right) x^{2}-24 s x+48-16 t^{2}=0
$$
$\Rightarrow x_{P}=\frac{12 s}{3 s^{2}+4 t^{2}}, y_{P}=\frac{12 t}{3 s^{2}+4 t^{2}}$. Substituting into $x_{P}^{2}+y_{P}^{2}=1 \Rightarrow\left(\frac{s^{2}}{4}+\frac{t^{2}}{3}\right)^{2}=s^{2}+t^{2}$, this is the equation of the trajectory of point $Q$. Clearly $|O Q|^{2}=s^{2}+t^{2}=\left(\frac{s^{2}}{4}+\frac{t^{2}}{3}\right)^{2} \in\left[\frac{1}{16}\left(s^{2}+t^{2}\right)^{2}, \frac{1}{9}\left(s^{2}+t^{2}\right)^{2}\right] \Rightarrow s^{2}+t^{2} \in[9,16]$, equality holds when $s=0$ or $t=0$. Therefore, the range of $|O Q|$ is $[3,4]$.