$$
\begin{array}{l}
A=\left\{(x, y) \mid x^{2}+y^{2}-10 x-16 y+81<0\right\}, \\
B=\{(x, y)|y \geqslant| x-t \mid+8\} .
\end{array}
$$
(1) If $A \cap B \neq \varnothing$, find the range of real number $t$;
(2) Suppose point $P(t, 8) \in A$, and the boundaries of the two planar regions represented by sets $A$ and $B$ intersect at points $M$ and $N$. Find the minimum value of $\frac{1}{|P M|}+\frac{1}{|P N|}$.
(1) Set
$$
A=\left\{(x, y) \mid(x-5)^{2}+(y-8)^{2}<(2 \sqrt{2})^{2}\right\} \text {. }
$$

When the ray $y=x-t+8(x \geqslant t)$ is tangent to the circle $(x-5)^{2}+(y-8)^{2}=(2 \sqrt{2})^{2}$, we have
$$
\frac{|5-8-t+8|}{\sqrt{2}}=2 \sqrt{2} \text {. }
$$

Solving this, we get $t=1$ or $t=9$ (discard).
When the ray $y=t-x+8(x \leqslant t)$ is tangent to the circle $(x-5)^{2}+(y-8)^{2}=(2 \sqrt{2})^{2}$, we have
$$
\frac{|5+8-t-8|}{\sqrt{2}}=2 \sqrt{2} \text {. }
$$

Solving this, we get $t=1$ (discard) or $t=9$.
Therefore, the range of $t$ is $1<t<9$.
(2) As shown in Figure 6, let the diameter of the circle represented by set $A$ parallel to the $x$-axis be $C D$, and the extension of $M P$ intersects the circle at another point $N^{\prime \prime}$. Then
$$
\begin{array}{c}
\frac{1}{|P M|}+\frac{1}{|P N|} \\
\geqslant \frac{2}{\sqrt{P M \cdot P N}} \\
\quad=\frac{2}{\sqrt{P M \cdot P N^{\prime}}} \\
\quad=\frac{2}{\sqrt{P C \cdot P D}} \\
\geqslant \frac{2}{\frac{P C+P D}{2}}=\frac{4}{C D}=\frac{4}{4 \sqrt{2}}=\frac{\sqrt{2}}{2} .
\end{array}
$$

When $t=5$, the equality holds.
Thus, the minimum value of $\frac{1}{|P M|}+\frac{1}{|P N|}$ is $\frac{\sqrt{2}}{2}$.