6. Given a function with period $T=4$
$$
f(x)=\left\{\begin{array}{ll}
m \sqrt{1-x^{2}}, & x \in(-1,1] ;(m>0) . \\
1-|x-2|, & x \in(1,3]
\end{array}\right.
$$

If the equation $3 f(x)=x$ has exactly five real solutions, then the range of values for $m$ is
6. $\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right)$.

To make the equation $3 f(x)=x$ have exactly five real solutions, the line $y=\frac{x}{3}$ must intersect with $y=m \sqrt{1-(x-4)^{2}}$ at two points, and not intersect with $y=m \sqrt{1-(x-8)^{2}}$ (as shown in Figure 3).
Figure 3
Solving the system of equations, for the former
$$
\Delta_{1}=4-\frac{20}{3 m^{2}}>0 \Rightarrow m^{2}>\frac{5}{3} \text {, }
$$

and for the latter
$$
\Delta_{2}=64-63\left(\frac{1}{9 m^{2}}+1\right)<0 \Rightarrow m^{2}<7 .
$$