11. It is known that,
$$
\begin{array}{l}
\triangle P O A \cong \triangle Q O B \\
\Rightarrow \angle P A O=\angle Q B O \\
\Rightarrow A, O, B, M \text{ are concyclic } \Rightarrow \angle A M B=90^{\circ} \\
\Rightarrow \angle P M Q=90^{\circ} .
\end{array}
$$

Therefore, the moving point $M$ lies on the circle with $P Q$ as its diameter. As shown in Figure 3.

According to the geometric conditions, point $M$ should be between the two tangents drawn from point $P$ to the unit circle, and between the two tangents drawn from point $Q$ to the unit circle.
Let $|M O|=d$. Then
$$
\cos \angle T_{1} M T_{2}=1-2 \sin ^{2} \angle O M T_{1}=1-\frac{2}{d^{2}} \text {. }
$$

Thus, $\overrightarrow{M T_{1}} \cdot \overrightarrow{M T_{2}}=\left|\overrightarrow{M T_{1}}\right|^{2} \cos \angle T_{1} M T_{2}$
$$
=\left(d^{2}-1\right)\left(1-\frac{2}{d^{2}}\right)=d^{2}+\frac{2}{d^{2}}-3 \text {. }
$$

Clearly, $d>1$.
When the moving point $M$ lies on the tangent line drawn from point $P$ to the unit circle, $|M O|$ is maximized, at this time,
$$
\begin{array}{l}
\angle M P O=30^{\circ},|M O|=\sqrt{2} \\
\Rightarrow 1<d^{2} \leqslant 2 .
\end{array}
$$

Note that, the function $x+\frac{2}{x}$ is decreasing in the interval $(1, \sqrt{2})$ and increasing in the interval $(\sqrt{2}, 2]$.
Thus, $2 \sqrt{2} \leqslant d^{2}+\frac{2}{d^{2}} \leqslant 3$.
Hence, the range of $\overrightarrow{M T_{1}} \cdot \overrightarrow{M T_{2}}$ is $[2 \sqrt{2}-3,0]$.
11. (20 points) As shown in Figure 1, $A$ and $B$ are two moving points on the circle $\odot O: x^{2}+y^{2}=1$, and the central angle corresponding to the arc $\overparen{A B}$ (from $A$ to $B$ in a counterclockwise direction) is $90^{\circ}$. Given points $P(2,0)$ and $Q(0,2)$, the intersection point of the lines $P A$ and $Q B$ is $M$, and $M$ is outside the circle $\odot O$. Two tangents are drawn from point $M$ to $\odot O$, with the points of tangency being $T_{1}$ and $T_{2}$. Find the range of $\overrightarrow{M T_{1}} \cdot \overrightarrow{M T_{2}}$.