(1)
-b<a<-a<b;
(2)∵-a+1>0,b-2<0,a-b<0, ∴|2(-a+1)|-|b-2|+2|a-b|, =2(-a+1)-[-(b-2)]+2[-(a-b)], =-4a+3b;
(3)|x+1|+|x-2|存在最小值,最小值为3. 当x<-1时,|x+1|+|x-2|=-x-1-x+2=-2x+1; 当-1≤x≤2时,|x+1|+|x-2|=x+1-x+2=3; 当x>2时,|x+1|+|x-2|=x+1+x-2=2x-1; ∴|x+1|+|x-2|存在最小值,最小值为3. |