$\because 3a=2b$，

$\therefore \dfrac{a}{b}=\dfrac{2}{3}$，

$\therefore$可设$a=2k$，那么$b=3k$，

$\therefore \dfrac{2a+3b}{2a-3b}=\dfrac{2\times 2k+3\times 3k}{2\times 2k-3\times 3k}=-\dfrac{13}{5}$.

故答案为$-\dfrac{13}{5}$.

$\because 3a=2b$，

$\therefore \dfrac{a}{b}=\dfrac{2}{3}$，

$\therefore$可设$a=2k$，那么$b=3k$，

$\therefore \dfrac{2a+3b}{2a-3b}=\dfrac{2\times 2k+3\times 3k}{2\times 2k-3\times 3k}=-\dfrac{13}{5}$.

故答案为$-\dfrac{13}{5}$.