7. Given that the domain of the function $f(x)$ is $(a, b)$, and $b-a>2$. Then the domain of $F(x)=f(3 x-1)-f(3 x+1)$ is ( ).
(A) $\left(\frac{a-1}{3}, \frac{b+1}{3}\right)$
(B) $\left(\frac{a+1}{3}, \frac{b-1}{3}\right)$
(C) $\left(\frac{a-1}{3}, \frac{b-1}{3}\right)$
(D) $\left(\frac{a+1}{3}, \frac{b+1}{3}\right)$
7.B.

From the problem, we have $\left\{\begin{array}{l}a2$, then
$$
\frac{b-1}{3}-\frac{a+1}{3}=\frac{(b-a)-2}{3}>0,
$$

which means $\frac{a+1}{3}<\frac{b-1}{3}$.
Therefore, the solution to the inequality system is $\frac{a+1}{3}<x<\frac{b-1}{3}$.
Hence, the domain of $F(x)$ is $\left(\frac{a+1}{3}, \frac{b-1}{3}\right)$.