(1)∵|a|=3,b2=4,∴a=±3,b=±2,∵ab<0,∴a=3,b=-2或a=-3,b=2,∴a-b=5或a-b=-5;(2)①∵A-2B=7a2-7ab,B=-4a2+6ab+7∴A=7a2-7ab+2B=7a2-7ab-8a2+12ab+14=a2+5ab+14;②∵|a+1|+(b-2)2=0,∴ a+1=0 b?2=0 ,解得 a=?1 b=2 ,∴A=a2+5ab+14=(-1)2+5×(-1)×2+14=3.