$解：(2)延长DP 交AB延长线于点 F$
$∵∠BAC+∠CDE=180°$
$∴AF//DE$
$∴∠PFB=∠PDE，∠PBF=∠PED$
$∵P 为BE的中点$
$∴BP=PE$
$在△BPF 和△EPD中$
$\begin{cases}{∠PFB=∠PDE}\\{∠PBF=∠PED}\\{PB=PE}\end{cases}$
$∴△BPF≌ △EPD (\mathrm {AAS})$
$∴BF=DE，PD=PF，S_{△PBF}=S_{△PDE}$
$∴S_{四边形ABED}=S_{△ADF}$
$∵DC=DE，∴DC=BF$
$∵AB=AC，AC∶CD=3∶5$
$∴AB∶BF=3∶5$
$∴S_{△ABP}∶S_{△BPF}=AB∶BF=3∶5$
$∵S_{△ABP}=6，∴S_{△BPF}=10，则S_{△APF}=16$
$∵PF=PD$
$∴S_{△ADP}=S_{△AFP}$
$∴S_{四边形ABED}=S_{△ADF}=2S_{△APF}=32$
