13. Let the inequality about $x$ be $\frac{x+2}{k}>1+\frac{x-3}{k^{2}}(k \in \mathbf{R}$ and $k \neq 0)$.
(1) Solve this inequality; (2) If the solution set of this inequality is $(3,+\infty)$, find the value of $k$; (3) If $x=3$ is a solution to the inequality, find the value of $k$.
13. (1) The original inequality is transformed into $(k-1) x>k^{2}-2 k-3$. When $k>1$, $x>\frac{k^{2}-2 k-3}{k-1}$; When $k-4, x \in \mathbf{R}$;
(2) $\frac{k^{2}-2 k-3}{k-1}=3$ (for $k>1$), we get $k=5$;
(3) Substituting $x=3$ gives $\frac{5}{k}>1$, we get $0<k<5$.