(1) In the first diagram, if $P \in AC$, we have $PA \cdot PC \le \frac{1}{4}$ and $PB \le \sqrt{2}$. Thus $PA \cdot PB \cdot PC \le \frac{\sqrt{2}}{4}$. The equality is not valid, since the two equality signs cannot be valid at the same time. Therefore $PA \cdot PB \cdot PC < \frac{\sqrt{2}}{4}$.

(2) In the second diagram, if $P \in AB$, write $AP = x \in [0, \sqrt{2}]$, then

Let $t = x(\sqrt{2} - x)$, then $t \in [0, \frac{1}{2}]$ and $f(x) = g(t) = t^2(1-t)$.

Note that $g'(t) = 2t - 3t^2 = t(2 - 3t)$. Thus $g(t)$ is increasing on $[0, \frac{2}{3}]$ and $f(x) \le g(\frac{1}{2}) = \frac{1}{8}$. Therefore $PA \cdot PB \cdot PC \le \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$. The equality is valid if and only if $t = \frac{1}{2}$ and $x = \frac{\sqrt{2}}{2}$. So $P$ is the midpoint of $AB$.