(1)f(x)=a·b
=(sinwx)^2+√3sinwxcoswx
=1/2+(√3/2sin2wx-1/2cos2wx)
=1/2+sin(2wx-π/6)
因T=2π/2w=π,即w=1
于是f(x)=1/2+sin(2x-π/6)
当π/2+2kπ≤2x-π/6≤3π/2+2kπ时f(x)递减
所以递减区间为[π/3+kπ,5π/6+kπ]

(2)因0≤x≤2π/3
则-π/6≤2x-π/6≤7π/6
即有-1/2≤sin(2x-π/6)≤1
所以0≤f(x)≤3/2