(1) f(x)=sin²ωx+2√3 sin(ωx+π/4)cos(ωx-π/4)-cos²ωx-√3
=2√3 ·√2/2(sinωx+cosωx)·√2/2(sinωx+cosωx)-(cos²ωx-sin²ωx)-√3
=√3(1+sin2ωx)-cos2ωx-√3
=√3sin2ωx-cos2ωx
=2sin(2ωx-π/6)
f[x+π/(6ω)]=2sin{2ω[x+π/(6ω)]-π/6}=2sin(2ωx+π/6)
一个单调增区间满足-π/2≤2ωx+π/6≦π/2
即-π/(3ω)≤x≦π/(6ω)
所以 -π/(3ω)≦-π/3 且 π/(6ω)≧2π/3
解得 0＜ω≦1/4
(2)设f(x)最小正周期为T,由题意,T/4=π/4,T=π
所以 2π/2ω=π, ω=1
所以f(x)=2sin(2x-π/6)
x∈[-π/12,25π/36]时,2x-π/6∈[-π/3,11π/9]
而 11π/9＜4π/3
所以f(x)∈[-√3,2]
当 2x-π/6=-π/3 即 x=-π/12 时,f(x)有最小值-√3；
当 2x-π/6=π/2 即 x=π/3 时,f(x)有最大值2.