$(2)解:過(guò)點(diǎn)P作PM⊥AD$
$∵菱形ABEF,∠ABC=60°$
$∴AB=AF=4,∠AFB=30°$
$∴AP=2$
$∴PF=2\sqrt{3}$
$S_{△APF}=\frac12×AP×PF=\frac12×AF×MP$
$MP=\sqrt{3}$
$∵∠AFB=60°$
$∴∠PAF=60°$
$在Rt△APM中$
$AM=\frac {\sqrt{3}}{3}×MP=1$
$MD=AD-AM=5$
$tan∠ADP=\frac {\sqrt{3}}{5}$
