$解:(1) 原式= 5a2-(3a-2a+3 + 4a2)$
$= 5a2-(a +3+ 4a2)$
$= 5a2-a-3-4a2$
$=a2-a- 3.$
$當(dāng)a=-2時,原式= (-2)2-(-2)-3= 4+2-3= 3.$
$(2)因為a2-b2=a2-ab+ab-b2= (a2-ab)+(ab-b2),$
$所以把a(bǔ)2-ab= 8,ab-b2=-4代入,得a2-b2= (a2-ab)+(ab-b2)=8+(-4)=4.$
$因為a2- 2ab+b2=a2-ab-ab+b2= (a2-ab)-(ab-b2),$
$所以把a(bǔ)2-ab= 8,ab-b2=-4代入,得a2-2ab+b2 = (a2-ab)- (ab-b2)=8-(-4) - 8+4= 12.$
