(1)将(0,0)代入g(x)中得：a=0
则：g(x)=bx-(1/2)x^2+(1/3)x^3
又由题意可知:在点(0,0)处f(x)与g(x)的切线斜率相同
则有：f'(0)=g'(0)
则：g'(0)=b-0+0^2=f'(0)=1/(1+0)
即:b=1
则：g(x)=x-(1/2)x^2+(1/3)x^3
(2)
设F(x)=g(x)-f(x)=x-(1/2)x^2+(1/3)x^3-ln(1+x)
则：F'(x)=1-x+x^2- 1/(1+x)
令F'(x)=0
则有：F'(x)=x^3/(1+x)=0
得：x=0
故当x>0时,x+1>0,F'(x)>0
则F(x)在[0,+∞)上单增
则：x>=0时,F(x)>=F(0)=0
即：g(x)>=f(x)