见解析
解:(1)取x1=x2=0
得f(0)≥f(0)+f(0),
又由f(0)≥0,得f(0)=0
(2)解:显然g(x)=2x-1在[0,1]上满足①g(x)≥0;②g(1)=1
若x1≥0,x2≥0,且x1+x2≤1,
则有g(x1+x2)-[g(x1)+g(x2)]=2x1+x2-1-[(2x1-1)+(2 x2-1)]=(2x1-1)(2 x2-1)≥0.
故g(x)=2x-1满足条件①﹑②﹑③
所以g(x)=2x-1为友谊函数.
(3)解:因为0≤x1<x2≤1,则0<x2-x1<1,
所以f(x2)=f(x2-x1+x1)≥f(x2-x1)+f(x1)≥f(x1)
故有f(x1)≤f(x2).