（1）f(x) = √3sinωxcosωx– cos2ωx + 3/2 = (√3/2)sin2ωx – (1/2)(1 + cos2ωx) + 3/2 = (√3/2)sin2ωx– (1/2)cos2ωx + 1 = sin(2ωx –π/6) + 1,函数f(x)的最小正周期为π,所以2π/|2ω| = π/|ω| = π,所以|ω| = 1,可得ω = 1或者-1,因为函数f(x)的图像关于x = π/6对称,所以f(π/6) = 1 + 1 = 2或者-1 + 1 = 0；
1）如果ω = 1,f(x) = sin(2x –π/6) + 1,此时f(π/6) = sin(π/6) + 1 = 3/2,舍去；
2）如果ω = -1,f(x) = sin(-2x– π/6) +1,此时f(π/6) = sin(-π/2) + 1 = 0,符合题意；
综上所述,函数f(x)的解析式为f(x) = sin(-2x – π/6) + 1；
（2）若关于 x的方程1 – f(x) = a在[0,π/2]上只有一个实数根,即-sin(-2x –π/6) = a,即sin(2x + π/6) = a在x∈[0,π/2]上只有一个实数根,记y1= sin(2x + π/6),其中x∈[0,π/2],记y2= a,由题意,这两个函数的图象有且只有一个公共点,数形结合可得a∈[-1/2,1/2)∪{1}.