(1)
f(x)=(2√3cosx+sinx)sinx-sin²(π/2+x)
=2√3sinxcosx+sin²x-(1/2)[1-cos(π+2x)]
=√3sin2x+(1/2)(1-cos2x)-1/2+(1/2)cos(π+2x)
=√3sin2x-(1/2)cos2x-(1/2)cos2x
=√3sin2x-cos2x
=2*[(√3/2)sin2x-(1/2)cos2x]
=2sin(2x-π/6)
所以：f(x)的最大值为2
单调递增区间满足：2kπ-π/2