f(x)=√3（sinωxcosωx)-(cosωx)^2+½
=√3/2sin2ωx-½ cos2ωx
=sin(2ωx-π/6)
据题T/2=π/4 ∴T=π/2 2π/2ω=π/2 ∴ω=2 f(X)=sin(4x-π/6)
(2)x∈（7π/24,5π/12）时,(4x-π/6)∈(π,3π/2)
f（x）=-3/5 ∴cos(4x-π/6)=-4/5
cos4x=cos(4x-π/6+π/6)
=√3/2 cos(4x-π/6)-sin(4x-π/6)/2
=√3/2 x (-4/5)-(-3/5)x½
=(3-4√3)/10
(3)据题cosx≥1/2,x∈（0,π） ∴x∈(0,π/3) ∴,(4x-π/6)∈(-π/6,7π/6)
又f（x）=m 有且仅有一个实根 ( 好像题有点问题哦!)
∴据图易知：m=1