Let the analytical expression of the quadratic function be
$$y=a(x-1)^{2}-2=a x^{2}-2 a x+a-2,$$

and $a>0, a \neq 2$, with the graph as shown in Figure 7.

The intersections of the graph with the $x$-axis and $y$-axis are $A\left(x_{1}, 0\right)$, $B\left(x_{2}, 0\right)$, and $C(0, a-2)$.
(1) From $|O C|^{2}=|O A| \cdot|O B|$, we have
$$\begin{array}{l} (a-2)^{2}=\left|x_{1} x_{2}\right| \\ =\left|\frac{a-2}{a}\right|, \end{array}$$

which simplifies to $a^{3}-4 a^{2}+4 a=|a-2|$.
When $0 < a < 2$, we have $a^{3}-4 a^{2}+5 a-2=0$. Thus, $(a-1)\left(a^{2}-3 a+2\right)=0$.

This gives $a_{1}=1$, $a_{2}=2$ (discard), and $a_{3}=1$ (discard).
When $a > 2$, we have $a^{3}-4 a^{2}+3 a+2=0$. Thus, $(a-2)\left(a^{2}-2 a-1\right)=0$.

This gives $a_{1}=2$ (discard), $a_{2}=1+\sqrt{2}$, and $a_{3}=1-\sqrt{2}<0$ (discard).
Therefore, $a=1+\sqrt{2}$,
$$y=(1+\sqrt{2}) x^{2}-(2+2 \sqrt{2}) x+\sqrt{2}-1.$$

Thus, the required function expression is
$$y=x^{2}-2 x-1$$

or $y=(1+\sqrt{2}) x^{2}-(2+2 \sqrt{2}) x+\sqrt{2}-1$.
(2) From $S_{\triangle A B C}=\frac{1}{2}|A B| \cdot|O C|$, we have the following two cases:

When $y=x^{2}-2 x-1$,
$|A B|=\left|x_{1}-x_{2}\right|$
$=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=2 \sqrt{2}$.
Also, $|O C|=1$, so $S_{\triangle A B C}=\frac{1}{2} \times 2 \sqrt{2} \times 1=\sqrt{2}$.
When $y=(1+\sqrt{2}) x^{2}-(2+2 \sqrt{2}) x+\sqrt{2}-1$ (graph not shown),
$$\begin{array}{l} |A B|=\left|x_{1}-x_{2}\right| \\ =\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=2 \sqrt{2(\sqrt{2}-1)} . \end{array}$$

Also, $|O C|=\sqrt{2}-1$, so
$$\begin{array}{l} S_{\triangle A B C}=\frac{1}{2} \times 2 \sqrt{2(\sqrt{2}-1)} \times(\sqrt{2}-1) \\ =(\sqrt{2}-1) \sqrt{2(\sqrt{2}-1)} . \end{array}$$

Therefore, the area of $\triangle A B C$ is $(\sqrt{2}-1) \sqrt{2(\sqrt{2}-1)}$ or $\sqrt{2}$.