(1)依题意,知1,b为方程ax2-3x+2=0的两根,且b>1,a>0∴a×12-3×1+2=0, a×b2-3×b+2=0解得,a=1,b=2(b=1舍去)(2)原不等式即为x2-3x+2<0,即 (x-1)(x-2)<0,∴1<x<2∴原不等式的解集为{x|1<x<2}.