降幂公式：sin²x=（1-cos2x）/2
cos²x=（1+cos2x）/2
sinxcosx=(sin2x)/2
(1)f(x)=sin²x+2√3 sinxcosx+3cos²x
=（1-cos2x）/2+√3sin2x+3*（1+cos2x）/2
=2+cos2x+√3sin2x
=2+2sin(2x+π/6)
令2kπ-π/2＜2x+π/6＜2kπ+π/2 得单调增区间是 kπ-π/3＜x＜kπ+π/6
令2kπ+π/2＜2x+π/6＜2kπ+3π/2 得单调减区间是 kπ+π/6＜x＜kπ+4π/3
（2）f(α)=3
即2+sin(2α+π/6)=3
sin(2α+π/6)=1
因为α∈(0,π)
所以 2α+π/6=π/2
2α+π/6=π/2
α=π/6