$解:(2)∠DME= 180°- 2∠A,证明如下:$
$在△ABC中,∠ABC+∠ACB=180-∠A$
$∵DM= ME= BM= MC$
$∴∠BMD+∠CME$
$=(180°-2∠ABC)+(180°-2∠ACB)$
$=360°-2(∠ABC+∠ACB)$
$= 360°-2(180° -∠A)$
$=2∠A$
$∴∠DME=180°-2∠A$
$(3)结论(1)成立,结论(2)不成立,理由如下:$
$在△ABC中,∠ABC+∠ACB= 180°- ∠BAC$
$∵DM=ME=BM=MC$
$∴∠BME+∠CMD=2∠ACB+2∠ABC$
$= 2(180°-∠BAC)= 360°-2∠BAC$
$∴∠DME=180°- (360°-2∠BAC)=2∠ BAC- 180°$