jmc

We work on the two parts of the given inequality separately. First, $3\le \frac{x}{2x-5}$ is equivalent to $$0\le \frac{x}{2x-5}-3=\frac{x-3(2x-5)}{2x-5}=\frac{-5x+15}{2x-5}.$$Making a sign table, we have: \begin{array}{c|cc|c} &$-5x+15$ &$2x-5$ &$\frac{-5x+15}{2x-5}$ \\ \hline$x<\frac{5}{2}$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$\frac{5}{2}<x<3$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$x>3$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]\end{array}Therefore, the inequality holds when $\frac{5}{2}<x<3,$ as well as the endpoint $x=3,$ which makes the right-hand side zero. The solution set to the first inequality is $(\frac{5}{2},3].$

Second, $\frac{x}{2x-5}<8$ is equivalent to $$\frac{x}{2x-5}-8=\frac{x-8(2x-5)}{2x-5}=\frac{-15x+40}{2x-5}<0.$$Making another sign table, we have: \begin{array}{c|cc|c} &$-15x+40$ &$2x-5$ &$\frac{-15x+40}{2x-5}$ \\ \hline$x<\frac{5}{2}$ &$+%%DISP_1%%amp;$-%%DISP_1%%amp;$-$\\ [.1cm]$\frac{5}{2}<x<\frac{8}{3}$ &$+%%DISP_1%%amp;$+%%DISP_1%%amp;$+$\\ [.1cm]$x>\frac{8}{3}$ &$-%%DISP_1%%amp;$+%%DISP_1%%amp;$-$\\ [.1cm]\end{array}It follows that the inequality holds when either $x<\frac{5}{2}$ or $x>\frac{8}{3}.$

The intersection of this solution set with $(\frac{5}{2},3]$ is $\boxed{(\frac{8}{3},3]},$ which is the solution set for both inequalities combined.