(1)令log2x=t,则t=-1,fmin=-8,配方可知，a/2=-1,b-a^2/2=-8,所以a=-2,b=-6(2)将a,b值带入即可因式分解(log2x+3)(log2x-1)>0,所以log2x<-3或log2x>1,从而，02
