22. Given that the function $f(x)$ is defined on $\mathbf{R}^{+}$, satisfies $f(x)$ is strictly decreasing on $\mathbf{R}^{+}$, and
$$
f(x)>\frac{1}{x^{2}},
$$

on $\mathbf{R}^{+}$, it always holds that $f^{2}(x) f\left(f(x)-\frac{1}{x^{2}}\right)=f^{3}(1)$.
(1) Find the function value $f(1)$;
(2) Provide a function that satisfies the conditions.
22. (1) When $x=1$, from (2) we get $f^{2}(1) f(f(1)-1)=f^{3}(1)$.

From (1), we have $f(1)>1$, so $f(f(1)-1)=f(1)$. Combining with the fact that $f(x)$ is strictly decreasing on $\mathbf{R}^{+}$, we know that $f(1)-1=1$, i.e., $f(1)=2$.
(2) Suppose $f(x)=\frac{a}{x^{2}}$, from (2) we get $\frac{a^{2}}{x^{4}} \cdot f\left(\frac{a}{x^{2}}-\frac{1}{x^{2}}\right)=2^{3}$, i.e., $\frac{a^{2}}{x^{4}} \frac{a}{\left(\frac{a}{x^{2}}-\frac{1}{x^{2}}\right)}=8$.
Simplifying, we get $a^{3}-8 a^{2}+16 a-8=0$ which solves to $a=2,3 \pm \sqrt{5}$.
Since $f(1)=2$, we have $a=2$, thus $f(x)=\frac{2}{x^{2}}$.
When $0<x_{1}<x_{2}$, we have $f\left(x_{1}\right)=\frac{2}{x_{1}^{2}}>\frac{2}{x_{2}^{2}}=f\left(x_{2}\right)$, so $f(x)$ is strictly decreasing. Also, $f(x)=\frac{2}{x^{2}}>\frac{1}{x^{2}}$, so $f(x)$ satisfies condition (1). On the other hand, $f^{2}(x) f\left(f(x)-\frac{1}{x^{2}}\right)=\left(\frac{2}{x^{2}}\right)^{2} \cdot f\left(\frac{2}{x^{2}}-\frac{1}{x^{2}}\right)=\frac{4}{x^{4}} \cdot f\left(\frac{1}{x^{2}}\right)=\frac{4}{x^{4}} \cdot \frac{2}{\left(\frac{1}{x^{2}}\right)^{2}}=8=f^{3}(1)$.
Therefore, $f(x)$ satisfies condition (2), so the function $f(x)=\frac{2}{x^{2}}$ meets both conditions given in the problem.