f(x)=sin²ωx＋√3cosωxsinωx
=(√3/2)sin2ωx－(1/2)cos2ωx＋(1/2)
=sin(2ωx－π/6)＋(1/2)
相邻两对称轴之间的距离是半个周期,则一个周期是π,得：2π/|2ω|=π,得：ω=1
即：f(x)=sin(2x－π/6)＋(1/2)

(1)f(π/6)=sin(π/3－π/6)＋(1/2)=(1＋√3)/2
(2)f(kπ＋π/12)=sin(2kπ)＋(1/2)
要使得其在[－π/6,π/3]上递增,则：其周期最小至少要2×(2π/3)=4π/3,
则：
2π/|2k|≥4π/3,得：0