①f(x)=sin^2wx+√3sinwxsin(wx+π/2)
=1/2(1-cos2wx) +√3sinwx coswx
=1/2(1-cos2wx) +√3/2sin2wx
=√3/2sin2wx-1/2cos2wx+1/2
=sin(2wx-π/6) +1/2
最小正周期为π
所以w=1,f(x)= sin(2x-π/6) +1/2.
所以函数值域是[-1/2,3/2].
②x∈[-π/12,π/2]时,f(x)= sin(2x-π/6) +1/2
则2x-π/6∈[-π/3,5π/6]
sin(2x-π/6)∈[-√3/2,1].
所以函数值域是[(-√3+1)/2,3/2].