20. Given that the function $f(x)$ is defined on $\mathbf{R}^{+}$ and satisfies the following two conditions:
(1) $f(x)$ is monotonically decreasing on $\mathbf{R}^{+}$ and $f(x)>\frac{1}{x^{2}}$;
(2) On $\mathbf{R}^{+}$, it always holds that $f^{2}(x) \cdot f\left[f(x)-\frac{1}{x^{2}}\right]=f^{3}(1)$,
(1) Find the value of $f(1)$;
(2) Provide a function $f(x)$ that satisfies the conditions and verify it.
20. (1) Since for any $x \in \mathbf{R}^{+} \cdot f^{2}(x) \cdot f\left[f(x)-\frac{1}{x^{2}}\right]=f^{3}(1)$, let $x=1$, we get $f^{2}(1) \cdot f[f(1)-1]=f^{3}(1)$. From the condition $f(x)>\frac{1}{x^{2}}, x \in \mathbf{R}^{-}$, we have $f(1)>1 \neq 0$, hence $f[f(1)-1]=f(1)$. Since $f(x)$ is monotonically decreasing on $\mathbf{R}^{-}$, we get $f(1)-1=1$, i.e., $f(1)=2$.
(2) Since $f(x)$ is a decreasing function on $\mathbf{R}^{+}$, and $f(x)>\frac{1}{x^{2}}$, we conjecture that $f(x)=\frac{2}{x^{2}}$ satisfies the conditions.
Verification (1) For any $00$,
$$

So $f(x)$ is a decreasing function on $\mathbf{R}$, and $f(x)>\frac{1}{x^{2}}$ is clearly true.
Verification (2): When $f(x)=\frac{2}{x^{2}}$,
$$
f^{2}(x) \cdot f\left[f(x)-\frac{1}{x^{2}}\right]=f^{2}(x) \cdot f\left(\frac{1}{x^{2}}\right)=\frac{4}{x^{4}} \cdot 2 x^{2}=8 ;
$$

and $f^{3}(1)=8$,
so $f^{2}(x) \cdot f\left[f(x)-\frac{1}{x^{2}}\right]=f^{3}(1)$.
In conclusion: The function $f(x)=\frac{2}{x^{2}}$ satisfies the conditions.