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

where $m>0$. If the equation $3 f(x)=x$ has exactly five real solutions, then the range of $m$ is ( ).
(A) $\left(\frac{4}{3}, \frac{8}{3}\right)$
(B) $\left(\frac{4}{3}, \sqrt{7}\right)$
(C) $\left(\frac{\sqrt{15}}{3}, \frac{8}{3}\right)$
(D) $\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right)$
12. D.

Notice that, $y=m \sqrt{1-x^{2}}$, which means $x^{2}+\frac{y^{2}}{m^{2}}=1\left(m^{2} \neq 1\right.$ when it is an ellipse $)$.
According to the problem, we have $\left\{\begin{array}{l}y=\frac{1}{3} x, \\ \frac{y^{2}}{m^{2}}+(x-4)^{2}=1,\end{array}\right.$
Substituting and ensuring $\Delta>0$, we get $m>\frac{\sqrt{15}}{3}$;
$$
\left\{\begin{array}{l}
y=\frac{1}{3} x, \\
\frac{y^{2}}{m^{2}}+(x-8)^{2}=1,
\end{array}\right.
$$

Substituting and ensuring $\Delta>0$, we get $m<\sqrt{7}$.
Therefore, the solution is $m \in\left(\frac{\sqrt{15}}{3}, \sqrt{7}\right)$.