(1)y=sin'x+2√3sinxcosx-cos'x
=√3*2sinxcosx-(cos'x-sin'x)
=√3sin2x-cos2x
=2(√3/2sin2x-1/2cos2x)
=2(cosπ/6sin2x-sinπ/6cos2x)
=2sin(2x-π/6)
∴T=2π/2=π
∴ymin=-2
(2)-π/2+2kπ≤2x-π/6≤π/2+2kπ
-π/6+2kπ≤2x≤2π/3+2kπ
-π/12+kπ≤x≤π/4+kπ
又0≤x≤π
所以增区间为x∈[0,π/4]∪[11/12π,π]