Five, the function $f(x)$ is defined on $\mathbf{R}^{+}$ and satisfies the following conditions:
(1) $f(x)$ is strictly decreasing on $\mathbf{R}^{+}$, and $f(x) > \frac{1}{x^{2}}$.
(2) On $\mathbf{R}^{+}$, it always holds that
$$
f^{2}(x) f\left(f(x)-\frac{1}{x^{2}}\right)=f^{3}(1) \text {. }
$$
(1) Find the function value $f(1)$;
(2) Provide a function $f(x)$ that satisfies the given conditions.
(1) When $x=1$, we get
$$
\left.f^{2}(1) f^{f} f(1)-1\right)=f^{3}(1) \text {. }
$$

Given (1) $f(1)>1$.
$$
\therefore f(f(1)-1)=f(1) \text {. }
$$
Since $f(x)$ is strictly decreasing on $\mathbf{R}^{+}$,
$$
\begin{array}{l}
\therefore f(1)-1=1 . \\
\therefore f(1)=2 .
\end{array}
$$
(2) Let $f(x)=\frac{a}{x^{2}}$. From (2) we get $\frac{a^{2}}{x^{4}} \cdot f\left(\frac{a}{x^{2}}-\frac{1}{x^{2}}\right)=2^{3}$, $\frac{a^{2}}{x^{4}} \cdot \frac{a}{\left(\frac{a}{x^{2}}-\frac{1}{x^{2}}\right)^{2}}=8$.
Simplifying, we get $a^{3}-8 a^{2}+16 a-8=0$.
Solving, we get $a=2,3 \pm \sqrt{5}$.
$$
\begin{array}{l}
\because f(1)=2, \\
\therefore a=2 . \therefore f(x)=\frac{2}{x^{2}} .
\end{array}
$$

When $0 < x_1 < x_2$,
$$
\begin{aligned}
f\left(x_{1}\right) & =\frac{2}{x_{1}^{2}}>\frac{2}{x_{2}^{2}}=f\left(x_{2}\right) .
\end{aligned}
$$

That is, $f\left(x_{1}\right)>f\left(x_{2}\right)$, so $f(x)$ is strictly decreasing.
Also, $f(x)=\frac{2}{x^{2}}>\frac{1}{x^{2}}$,
$\therefore f(x)$ satisfies condition (1).
Since
$$
\begin{aligned}
f^{2} & (x) f\left(f(x)-\frac{1}{x^{2}}\right) \\
& =\left(\frac{2}{x^{2}}\right)^{2} \cdot f\left(\frac{2}{x^{2}}-\frac{1}{x^{2}}\right) \\
& =\frac{4}{x^{4}} \cdot f\left(\frac{1}{x^{2}}\right)=\frac{4}{x^{4}} \cdot \frac{2}{\left(\frac{1}{x^{2}}\right)^{2}} \\
& =8=f^{3}(1) .
\end{aligned}
$$

That is, $f(x)$ satisfies condition (2).
Therefore, the function $f(x)=\frac{2}{x^{2}}$ satisfies both conditions given in the problem.