8. Given the sequence $a_{n}$ satisfies
$$
\begin{array}{l}
a_{1}=m\left(m \in \mathbf{N}_{+}\right), \\
a_{n+1}=\left\{\begin{array}{ll}
\frac{a_{n}}{2}, & a_{n} \text { is even; } \\
3 a_{n}+1, & a_{n} \text { is odd. }
\end{array}\right.
\end{array}
$$

If $a_{4}=7$, then the possible values of $m$ are
8.9 or 56.
(1) $m$ is an odd number.

From $a_{2}=3 a_{1}+1=3 m+1$ being even, we know $a_{3}=\frac{3 m+1}{2}$, $a_{4}=\frac{3 m+1}{4}=7$ or $a_{4}=\frac{3(3 m+1)}{2}+1=7$.
Solving gives $m=9$ or 1.
Upon verification, $m=9$.
(2) $m$ is an even number.
(i) When $m=4 m_{0}$,
$$
\begin{array}{l}
a_{2}=\frac{a_{1}}{2}=2 m_{0}, a_{3}=\frac{a_{2}}{2}=m_{0}, \\
a_{4}=\frac{m_{0}}{2}=7 \text { or } a_{4}=3 m_{0}+1=7 .
\end{array}
$$

Solving gives $m_{0}=14$ or $2 \Rightarrow m=56$ or 8.
Upon verification, $m=56$.
$$
\begin{array}{l}
\text { (ii) When } m=4 m_{0}+2 \text {, } \\
a_{2}=\frac{a_{1}}{2}=2 m_{0}+1, \\
a_{3}=3\left(2 m_{0}+1\right)+1=2\left(3 m_{0}+2\right) .
\end{array}
$$

Thus, $a_{4}=3 m_{0}+2=7$ has no solution.
In summary, $m=9$ or 56.