解:$ (1) $因为$\frac {AD}{BD}=\frac {3}{2},$
设$AD=3k,$$BD=2k,$
则$AB=AD+BD=3k+2k=5k$
$ $所以$\frac {AB}{BD}=\frac {5k}{2k}=\frac {5}{2}$
$ (2) $因为$\frac {AE}{EC}=\frac {3}{2},$
设$AE=3k,$$EC=2k,$
则$AC=AE+EC=3k+2k=5k$
$ $所以$\frac {EC}{AC}=\frac {2k}{5k}=\frac {2}{5}$