f(x)=4sinxcos（x+π/3)+根号3
=4sinx(1/2cosx-√3/2sinx)+√3
=2sinxcosx-2√3sin²x+√3
=sin2x-2√3sin²x+√3(sin²x+cos²x)
=sin2x+√3cos²x-√3sin²x
=sin2x+√3(cos²x-sin²x)
=sin2x+√3con2x
=2sin(2x+π/3)
（1）最小正周期T=2π/2=π
-π/2+2Kπ≤2x+π/3≤π/2+2Kπ
解得x∈[-5π/12+kπ,π/12+kπ],k∈{0,1,2.}
(2)x∈[-π/3,π/4] 则2x+π/3∈[-π/3,5π/6]
由图像知在2x+π/3=-π/3时取得最小值=-√3
2x+π/3=π/2取得最大值=2