由丨x1-x2丨最小值為π/2得：T/2=π/2,所以T=π,因此ω=1,即f(x)=sin(2x+π/6)
(1)由-π/2+2kπ≤2x+π/6≤π/2+2kπ得：-π/3+kπ≤x≤π/6+kπ,即函數(shù)f(x)的單調(diào)增區(qū)間為
[-π/3+kπ,π/6+kπ](k∈z)
(2)因為f(x)=1/3,α∈[-π/3,π/6],所以f(α=π/6)=1/3
(3)f(x+π/6)+mcosx+3=0即sin(2x+π/2)+mcosx+3=0,所以cos2x+mcosx+3=0,即
2cos^2x+mcosx+2=0,
因為方程f(x+π/6)+mcosx+3=0在x∈（0,π/2）有實數(shù)解,所以有
△≥0
x1+x2>0
解這個不等式組可得：m≤-4