(12) (20 points) Given the ellipse $\frac{x^{2}}{4}+y^{2}=1$, and $P$ is any point on the circle $x^{2}+y^{2}=16$. Draw the tangents $P A$ and $P B$ to the ellipse from $P$, with the points of tangency being $A$ and $B$, respectively. Find the maximum and minimum values of $\overrightarrow{P A} \cdot \overrightarrow{P B}$.
(12) Let point $P(m, n), A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then the tangent lines $P A: \frac{x_{1} x}{4} + y_{1} y=1, P B: \frac{x_{2} x}{4} + y_{2} y=1$, because the tangent lines $P A$ and $P B$ both pass through point $P$, so we have $\frac{x_{1} m}{4} + y_{1} n=1$ and $\frac{x_{2} m}{4} + y_{2} n=1$ simultaneously, thus the equation of line $A B$ is: $\frac{m x}{4} + n y=1$.
Combining the equations of line $A B$ and the ellipse, we get:
$$
\left\{\begin{array}{l}
\frac{x^{2}}{4} + y^{2}=1, \\
\frac{m}{4} x + n y=1,
\end{array}\right.
$$

Eliminating $y$ gives:
$$
\left(4 n^{2} + m^{2}\right) x^{2} - 8 m x + \left(16 - 16 n^{2}\right)=0,
$$

Thus,
$$
\begin{aligned}
x_{1} + x_{2} & =\frac{8 m}{4 n^{2} + m^{2}}, \quad x_{1} x_{2}=\frac{16 - 16 n^{2}}{4 n^{2} + m^{2}}, \\
\Delta & =64 m^{2} - 4\left(4 n^{2} + m^{2}\right)\left(16 - 16 n^{2}\right) \\
& =64 n^{2}\left(m^{2} + 4 n^{2} - 4\right).
\end{aligned}
$$

From $\frac{m x}{4} + n y=1$, we get:
$$
\begin{array}{l}
y_{1} y_{2}=\frac{4 - m x_{1}}{4 n} \cdot \frac{4 - m x_{2}}{4 n}=\frac{4 - m^{2}}{\left(4 n^{2} + m^{2}\right)}, \\
\left(y_{1} + y_{2}\right) n=2 - \frac{m\left(x_{1} + x_{2}\right)}{4}=\frac{8 n^{2}}{4 n^{2} + m^{2}},
\end{array}
$$

Thus,
$$
\begin{aligned}
\overrightarrow{P A} \cdot \overrightarrow{P B} & =\left(x_{1} - m, y_{1} - n\right) \cdot\left(x_{2} - m, y_{2} - n\right) \\
& =x_{1} x_{2} - \left(x_{1} + x_{2}\right) m + m^{2} + y_{1} y_{2} - \left(y_{1} + y_{2}\right) n + n^{2} \\
& =\frac{20 - 3 m^{2}}{4 n^{2} + m^{2}} + m^{2} + n^{2} - 6.
\end{aligned}
$$

Since $m^{2} + n^{2}=16$, we have $m^{2}=16 - n^{2}$, substituting this into the above equation gives:
$$
\overrightarrow{P A} \cdot \overrightarrow{P B}=11 - \frac{44}{3 n^{2} + 16},
$$

Since $0 \leqslant n^{2} \leqslant 16$, when $n^{2}=0, m^{2}=16$, i.e., point $P( \pm 4,0)$, $\overrightarrow{P A} \cdot \overrightarrow{P B}$ has a minimum value of $\frac{33}{4}$, and when $n^{2}=16, m^{2}=0$, i.e., point $P(0, \pm 4)$, $\overrightarrow{P A} \cdot \overrightarrow{P B}$ has a maximum value of $\frac{165}{16}$.