答：
f(x)=2sin(x+π/2)sin(x+7π/3)-√3sin²x+sin(x+π)cos(x+3π)
=-cos(2x+π/2+7π/3)+cos(7π/3-π/2)+(√3/2)(cos2x-1)+(-sinx)*(-cosx)
=-cos(2x+5π/6)+cos(π/6)+(√3/2)cos2x-√3/2+(1/2)sin2x
=-cos(2x+5π/6)+cos(2x-π/6)
=-2sin[(2x-π/6+2x+5π/6)/2] * sin[(-π/6-5π/6)/2]
=2sin(2x+π/3)
1)
f(x)的单调递增区间满足：2kπ-π/2