（1）
f(x)=sin²wx+(根号3/2)sin2wx-(1/2)
=1/2(1-cos2wx)+√3/2sin2wx-1/2
=√3/2sin2wx-1/2cos2wx
=sin(2wx-π/6)
∵f(x)的最小正周期为2π
∴2π/(2w)=2π
∴w=1/2
∴f(x)=sin(x-π/6)
由2kπ-π/2≤x-π/6≤2kπ+π/2,k∈Z
得2kπ-π/3≤x≤2kπ+2π/3,k∈Z
∴f(x)的单调递增区间是
[2kπ-π/3,2kπ+2π/3],k∈Z
(2)
∵x∈【-（π/6）,（5π/6）】
∴x-π/6∈[-π/3,2π/3]
∴x-π/6=-π/3,x-π/6时,f(x)取得最小值-√3/2
x-π/6=π/2,x=2π/3时,f(x)取得最大值1