13. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and satisfies the following two conditions: (1) For any $x, y \in \mathbf{R}$, $f(x+y)=f(x)+f(y)$; (2) When $x>0$, $f(x)<0$, and $f(1)=-2$. Find the maximum and minimum values of the function $f(x)$ on $[-3,3]$.
13. First, determine the monotonicity of $f(x)$ on $[-3,3]$. Let $0 \leqslant x_{1} \leqslant x_{2} \leqslant 3$, then $f\left(x_{2}\right)=f\left[\left(x_{2}-x_{1}\right)+x_{1}\right]=$ $f\left(x_{2}-x_{1}\right)+f\left(x_{1}\right), f\left(x_{2}\right)-f\left(x_{1}\right)=f\left(x_{2}-x_{1}\right)$. Since $x_{2}-x_{1}>0, f\left(x_{2}-x_{1}\right)<0$. Therefore, $f\left(x_{2}\right)<$ $f\left(x_{1}\right)$. Hence, $f(x)$ is a decreasing function on $[0,3]$. Since $f(x)$ is an odd function, $f(x)$ is also a decreasing function on $[-3,0]$. Thus, $f(x)$ is a decreasing function on $[-3,3]$, so $f_{\max }(x)=f(-3)=-3 f(1)=6 . f_{\min }(x)=f(3)=-6$.