1. Given $f(x)=\frac{\sqrt{1-x^{2}}}{|x+2|-2}, g(x)=$ $\lg \left(\sqrt{1+x^{2}}-x\right)$. Then the product function $F(x)=f(x) g(x)$ on the common domain is ()
(A) is an odd function but not an even function
(C) is both an odd function and an even function
(I)) is neither an odd function nor an even function
$-1 .(\mathrm{B})$.
First, find the domain of the two functions. For $f(x)$, we have
$$
\left\{\begin{array}{l}
1-x^{2} \geqslant 0, \\
|x+2|-2 \neq 0 .
\end{array}\right.
$$

This gives $\left\{\begin{array}{l}-1 \leqslant x \leqslant 1, \\ x \neq 0 \text { and } x \neq-4 .\end{array}\right.$
Thus, the domain is $(-1,0) \cup(0,1)$.
The domain of $g(x)$ is $\mathbf{R}$, so the common domain of the product function is $(-1,0) \cup(0,1)$.
For $x \in(-1,0) \cup(0,1)$, we have
$$
|x+2|-2=2+x-2=x \text {. }
$$

Thus, $f(x)=\frac{\sqrt{1-x^{2}}}{x}$.
We have $f(-x)=-f(x)$.
$$
\begin{aligned}
\text { Also, } & g(-x)+g(x) \\
& =\lg \left(\sqrt{1+x^{2}}+x\right)+\lg \left(\sqrt{1+x^{2}}-x\right) \\
& =\lg \left(\sqrt{1+x^{2}}+x\right)\left(\sqrt{1+x^{2}}-x\right) \\
& =\lg 1=0,
\end{aligned}
$$

Thus, $g(-x)=-g(x)$.
We get
$$
\begin{aligned}
F(-x) & =f(-x) g(-x) \\
& =[-f(x)][-g(x)] \\
& =f(x) g(x)=F(x) .
\end{aligned}
$$

By definition, $F(x)$ is an even function on $(-1,0) \cup(0,1)$. Since $F(x)$ is not identically zero, it cannot be an odd function. Therefore, the correct choice is (B).