(2) The domain of the function $f(x)$ is $\mathbf{R}$. If $f(x+1)$ and $f(x-1)$ are both odd functions, then ( ).
(A) $f(x)$ is an even function
(B) $f(x)$ is an odd function
(C) $f(x+3)$ is an odd function
(D) $f(x+3)$ is an even function
(2) C Hint: Since $f(x+1)$ is an odd function, we have
$$
f(x+1)=-f(-x+1), f(x)=-f(-x+2) \text{. }
$$

Similarly, since $f(x-1)$ is an odd function, we get
$$
f(x)=-f(-x-2).
$$

From (1) and (2), we can deduce that $f(-x+2)=f(-x-2)$, so $f(x+2)=$ $f(x-2), f(x+4)=f(x)$, which means $f(x)$ is a periodic function with a period of 4.
Since $f(x-1)$ is an odd function, $f(x+3)$ is also an odd function. Therefore, the answer is C.