解：（1）$\because AB$是$\odot O$的直径，弦$CD\perp AB，$$CD = 16，$
$\therefore \angle OED = 90^{\circ}，$$DE=\frac{1}{2}CD = 8。$
$\because BE = 4，$$\therefore OE = OB - BE = OD - 4。$
在$Rt\triangle OED$中，$\because OE^{2}+DE^{2}=OD^{2}，$
$\therefore (OD - 4)^{2}+8^{2}=OD^{2}，$
$OD^{2}-8OD + 16+64=OD^{2}，$
$-8OD=-80，$
$\therefore OD = 10。$
$\therefore \odot O$的直径是$20。$
（2）由（1），得$\angle OED = 90^{\circ}，$$\therefore \angle EOD+\angle D = 90^{\circ}。$
$\because \angle M=\angle D，$$\angle EOD = 2\angle M，$
$\therefore \angle EOD+\angle D = 2\angle M+\angle D = 2\angle D+\angle D = 90^{\circ}，$
$3\angle D = 90^{\circ}，$
$\therefore \angle D = 30^{\circ}。$
