首先更正一下题目：原题应为：f(x)=sin^2ωx+√3sinωxsin(ωx+π/2) (ω＞0).
f(x)=（1-cos2ωx)/2+√3sinωxcosωx.
=(√3/2)sin2ωx-(1/2)cos2ωx+1/2.
=sin(2ωx-π/6)+1/2.
∵T=2π/2ω=π.∴ω=1.
∴f(x)=sin(2x-π/6)+1/2.
当sin(2x-π/6)=1,x=5π/12时,sin(2x-π/6)取得最大值1,函数f(x)取得最大值,f(x)max=1+1/2=3/2.
当sin(2x-π/6)=-1,x=11π/12时,sin(2x-π/6)取得最小值-1,函数f(x)取得最小值：f(x)min=-1+1/2=-1/2.
∴函数f(x)=sin(2x-π/6)在区间[0,2π/3]上的取值范围为[-1/2,3/2].【[5π/12,11π/12]∈[0,2π/3]】