5. Let $f(x)$ and $g(x)$ be odd and even functions defined on $\mathbf{R}$, respectively. When $x>0$, and $g(3)=0$. Then the solution set of the inequality $f(x) \cdot g(x)<0$ is
5. $(-\infty,-3) \cup(0,3)$ Let $H(x)=f(x) \cdot g(x)$, from the given we know that $H(x)$ is an odd function, and $H^{\prime}(x)=$ $f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$. Therefore, when $x<0$, knowing that $H^{\prime}(x)>0$, $H(x)$ is monotonically increasing on $(-\infty, 0)$; since $H(x)$ is an odd function, $H(x)$ is also monotonically increasing on $(0,+\infty)$, and $H(3)=-H(-3)=0$. Based on the above information, we can draw the schematic diagram of $H(x)$ (omitted), from which we can obtain that the solution set of the inequality $f(x) \cdot g(x)<0$ is $(-\infty,-3) \cup(0,3)$.