7. 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 $\qquad$ .
7. Let \( H(x) = f(x) \cdot g(x) \). From the given conditions, \( H(x) \) is an odd function. Also, \( H'(x) = f'(x) g(x) + f(x) g'(x) \). Therefore, when \( x < 0 \), we know that \( H(x) \) is increasing on \( (-\infty, 0) \). Since \( H(x) \) is an odd function, \( H(x) \) is also increasing on \( (0, +\infty) \), and \( H(3) = -H(-3) = 0 \). Based on the above information, we can sketch the graph of \( H(x) \) (omitted). From the graph, we can determine that the solution set of the inequality \( f(x) \cdot g(x) < 0 \) is \( (-\infty, -3) \cup (0, 3) \).