5. Let the even function $f(x)$ satisfy: $f(1)=2$, and when $x y \neq 0$, $f\left(\sqrt{x^{2}+y^{2}}\right)=\frac{f(x) f(y)}{f(x)+f(y)}$. Then $f(5)=$ $\qquad$.
5. $\frac{2}{25}$.

From the condition, we have
$$
\frac{1}{f\left(\sqrt{x^{2}+y^{2}}\right)}=\frac{1}{f(x)}+\frac{1}{f(y)} .
$$

Since $f(-x)=f(x)$, we only need to consider the case where $x, y \in \mathbf{R}$.
Let $x=\sqrt{u}, y=\sqrt{v}$, then
$$
\frac{1}{f(\sqrt{u+v})}=\frac{1}{f(\sqrt{u})}+\frac{1}{f(\sqrt{v})} .
$$

Define
$$
g(t)=\left\{\begin{array}{ll}
\frac{1}{f(\sqrt{t})}, & f(\sqrt{t}) \neq 0 \\
0, & f(\sqrt{t})=0 .
\end{array}\right.
$$

Then it becomes
$$
g(u+v)=g(u)+g(v) .
$$

From equation (1), we know
$$
\begin{array}{l}
g(2 u)=g(u+u)=2 g(u) \\
g(3 u)=g(u+2 u)=g(u)+2 g(u)=3 g(u)
\end{array}
$$

By induction, we get $g(k u)=k g(u)$.
Let $k=25, u=1$, then $g(25)=25 g(1)$, i.e.,
$$
\frac{1}{f(\sqrt{25})}=\frac{25}{f(\sqrt{1})} \Rightarrow f(5)=\frac{2}{25} \text {. }
$$