5. If $a>0, a \neq 1, F(x)$ is an odd function, then $G(x)=F(x) \cdot\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right)$ is
(A) an odd function;
(B) an even function;
(C) neither an odd function nor an even function;
(D) the parity depends on the specific value of $a$.
5. (B)
$\because \quad F(x)$ is an odd function, $\quad \therefore \quad F(-x)=-F(x)$
Let $\quad g(x)=\frac{1}{a^{x}-1}+\frac{1}{2}=\frac{a^{x}+1}{2\left(a^{x}-1\right)}$,

then
$$
\begin{aligned}
g(-x) & =\frac{a^{-x}+1}{2\left(a^{-x}-1\right)}=\frac{\frac{1}{a^{x}}+1}{2\left(\frac{1}{a^{x}}-1\right)} \\
& =\frac{a^{x}+1}{2\left(1-a^{x}\right)}=-g(x),
\end{aligned}
$$

i.e., $g(x)$ is also an odd function.
When $a>0, a \neq 1$, $G(x)=F(x) g(x)$ is defined on $R$.
And $G(-x)=F(-x) g(-x)=[-F(x)][-g(x)]$
$$
=F(x) g(x),
$$
$\therefore G(x)$ is an even function.