当x≥1时,|x+3|-|x-1|=x+3-(x-1)=4;
当-3≤x＜1时,|x+3|-|x-1|=x+3-(1-x)=2x+2＜4;
当x＜3时,|x+3|-|x-1|=-x-3-(1-x)=-4;
所以,|x+3|-|x-1|≤4恒成立;
据题意可得a^2-3a＞4;
所以a＞4或a＜-1.