(2)解:$\because EP$平分$∠ AEH$,
$\therefore∠ AEH=2α=40°.$
$\because AB// CD$,
$\therefore∠ END=∠ AEH=40°.$
又$\because FG$平分$∠ DFI$,
$\therefore∠ IFG=∠ DFG=β=50°$,
$\therefore∠ CFI=180°-2β=80°.$
(3)解:$\because FG$平分$∠ DFI$,
$\therefore∠ DFG=∠ IFG=∠ PFC=β$,
$\therefore∠ CFI=180°-2β.$
$\because EP$平分$∠ AEH$,
$\therefore∠ AEH=2∠ AEP=2α.$
$\because AB// CD$,
$\therefore∠ AEN=∠ END=2α$,
$\therefore$当$FI// EH$时,$∠ END=∠ CFI$,
即$2α=180°-2β$,
$\thereforeα+β=90°.$
