(1)若【存在】a∈[1,3]使不等式ax²+(a-2)x-2＞0成立,(a>0)故ax²+(a-2)x-2＞0得(ax-2)(x+1)＞0
x>2/a或x<-1
(2)同(1)
(3)由ax²+(a-2)x-2＞0得(ax-2)(x+1)＞0,若不存在a∈[1,3]使不等式ax²+(a-2)x-2＞0成立,即a∈[1,3]使不等式ax²+(a-2)x-2<=0成立,由ax²+(a-2)x-2<=0得:(ax-2)(x+1)<=0即:2/a>=x>=-1