答：
f(x)=|x+1|-|x|+a
1）a=0
f(x)=|x+1|-|x|>=0
|x+1|>=|x|
两边平方得：
x^2+2x+1>=x^2
解得：x>=-1/2
2）
f(x)=|x+1|-|x|+a<=2恒成立
x<=-1时：-x-1+x+a<=2,a<=3
-1<=x<=0时：x+1+x+a<=2,-2<=2x<=1-a<=0,-3<=-a<=-1,1<=a<=3
x>=0时：x+1-x+a<=2,a<=1
综上所述,a<=1