(1) Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$. Draw perpendiculars from point $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$, respectively. Then the value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$.
(1) -1 Hint: Method one Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$, then the equation of line $P A$ is $y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right)$, i.e., $y=-x+2 x_{0}+\frac{2}{x_{0}}$.
From $\left\{\begin{array}{l}y=x, \\ y=-x+2 x_{0}+\frac{2}{x_{0}}\end{array}\right.$, we get $A\left(x_{0}+\frac{1}{x_{0}}, x_{0}+\frac{1}{x_{0}}\right)$.
Also, $B\left(0, x_{0}+\frac{2}{x_{0}}\right)$, so $\overrightarrow{P A}=\left(\frac{1}{x_{0}},-\frac{1}{x_{0}}\right), \overrightarrow{P B}=\left(-x_{0}, 0\right)$. Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=\frac{1}{x_{0}} \cdot\left(-x_{0}\right)=-1$.
Method two As shown in the figure, let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$ $\left(x_{0}>0\right)$, then the distances from point $P$ to the line $x-y=0$ and the $y$-axis are respectively
$|P A|=\frac{\left|x_{0}-\left(x_{0}+\frac{2}{x_{0}}\right)\right|}{\sqrt{2}}=\frac{\sqrt{2}}{x_{0}}$,
$|P B|=x_{0}$.
Since $O, A, P, B$ are concyclic (with $O$ being the origin), we have $\angle A P B=$ $\pi-\angle A O B=\frac{3 \pi}{4}$.
Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=|\overrightarrow{P A}| \cdot|\overrightarrow{P B}| \cos \frac{3 \pi}{4}=-1$.