(13 Given $P$ is a moving point on the circle $(x-2)^{2}+(y-2)^{2}=1$, vector $\overrightarrow{O P}$ is rotated $90^{\circ}$ counterclockwise to get vector $\overrightarrow{O S}$, and the point $P$ is symmetric to point $T$ with respect to $A(3,0)$. Find the range of $|\overrightarrow{T S}|$.
13 Let point $P(x, y)$, then point $S(-y, x)$, point $T(6-x, -y)$. The center of the circle is $C(2,2)$, and the radius $r=1$.
So
$$
\begin{aligned}
\overrightarrow{|T S|} & =\sqrt{(6-x+y)^{2}+(-y-x)^{2}} \\
& =\sqrt{2 x^{2}+2 y^{2}-12 x+12 y+36} \\
& =\sqrt{2} \cdot \sqrt{(x-3)^{2}+(y+3)^{2}}
\end{aligned}
$$
$$
=\sqrt{2}|P B| \text {, (where point } B(3,-3) \text { ) }
$$

Therefore, $\left.\overrightarrow{\mid S S}\right|_{\text {max }}=\sqrt{2}(|C B|+1)=\sqrt{2}(\sqrt{26}+1)$,
$$
\left.\overrightarrow{\mid T S}\right|_{\text {min }}=\sqrt{2}(|C B|-1)=\sqrt{2}(\sqrt{26}-1) \text {. }
$$

Thus, $\sqrt{2}(\sqrt{26}-1) \leqslant|\overrightarrow{\mid T S}| \leqslant \sqrt{2}(\sqrt{26}+1)$.