1. The function $f(x)$ defined on $\mathbf{R}$ satisfies
$$
f(x+3)+f(x)=0,
$$

and the function $f\left(x-\frac{3}{2}\right)$ is an odd function. Given the following 3 propositions:
(1) The period of the function $f(x)$ is 6;
(2) The graph of the function $f(x)$ is symmetric about the point $\left(-\frac{3}{2}, 0\right)$;
(3) The graph of the function $f(x)$ is symmetric about the $y$-axis. Among them, the number of true propositions is ( ).
(A) 3
(B) 2
(C) 1
(D) 0
$-1 . \mathrm{A}$.
From $f(x+3)=-f(x)$, we know $f(x+6)=f(x)$. Therefore, (1) is correct; Shifting the graph of $f\left(x-\frac{3}{2}\right)$ to the left by $\frac{3}{2}$ units results in the graph of $f(x)$, and the graph of $f\left(x-\frac{3}{2}\right)$ is symmetric about the origin, so, (2) is correct; From (2), we know $f(-x)=-f(-3+x)=$ $f(x)$, thus $f(x)$ is an even function, so, (3) is correct.