5. Given that the even function $f(x)$ satisfies for any $x \in$ $\mathbf{R}$,
$$
f(1+x)=f(3-x),
$$

and $f(x)=\left\{\begin{array}{ll}m \sqrt{1-x^{2}}, & x \in[0,1] ; \\ x-1, & x \in(1,2] \text {; }\end{array}\right.$
If the equation $3 f(x)=x$ has exactly 5 real solutions, then the range of the real number $m$ is
5. $\left(-\sqrt{7},-\frac{\sqrt{15}}{3}\right) \cup\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right)$.

It is easy to see that the function $f(x)$ is a periodic function with a period of 4.
We discuss three cases for $m$.
(1) When $m=0$, the original equation has exactly three distinct real roots, which does not meet the requirement.
(2) When $m>0$, by the graphs of the function $f(x)$ and the function $y=\frac{x}{3}$, we know that the line $y=\frac{x}{3}$ intersects the graph of
$$
y=m \sqrt{1-(x-4)^{2}}(x \in(3,5])
$$

at two different points, and does not intersect the graph of
$$
y=m \sqrt{1-(x-8)^{2}}(x \in(7,9])
$$

Thus, $\frac{\sqrt{15}}{3}<m<\sqrt{7}$.
(3) When $m<0$, similarly, we find that
$$
-\sqrt{7}<m<-\frac{\sqrt{15}}{3} \text {. }
$$

In summary, the range of $m$ is
$$
\left(-\sqrt{7},-\frac{\sqrt{15}}{3}\right) \cup\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right) .
$$