f(x)=(1/3)x^3+ax+b,
f'(x)=x^2+a,
f(x)在x=2处取最小值-4/3,
∴f'(2)=4+a=0,a=-4,
f(2)=8/3-8+b=-4/3,b=4,
∴f(x)=(1/3)x^3-4x+4<=m^2+m+103在[-4,3]上恒成立,
f'(x)=x^2-4=(x+2)(x-2),-2f(-2)=-8/3+12=28/3,f(3)=1,
∴f(x)|max=28/3<=m^2+m+103,
∴m^2+m+281/3>=0,
∴m的取值范围是R.为什么m的取值范围是Rm^2+m+281/3=(m+1/2)^2+281/3-1/4>0,