解：f(x)=2cosxsin(x+π/3)-√3(sinx)^2+sinxcosx
=2cosx(sinxcosπ/3+cosxsinπ/3)-√3(sinx)^2+sinxcosx
=2cosx(sinx1/2+cosx√3/2)-)-√3(sinx)^2+sinxcosx
=cosxsinx+√3(cosx)^2)-√3(sinx)^2+sinxcosx
=2cosxsinx+√3(cosx)^2)-√3(sinx)^2
=sin2x+√3cos2x
=2[(1/2)sin2x+(√3/2)cos2x]
=2[cosπ/3 sin2x+sinπ/3 cos2x]
=2sin(2x+π/3)
（1）最小正周期是T=2π/2=π
（2）函数最大值f(x)=2,最小值f(x)=-2
（3）
-π/2+2kπ≤2x+π/3≤π/2+2kπ
-5π/6+2kπ≤2x≤π/6+2kπ
-5π/12+kπ≤x≤π/12+kπ
函数的递增区间是[-5π/12+kπ，π/12+kπ]