（1）
T=2π/w=π
解得w=2
f(x)=√3cos2x
f(a)=√3cos2a=√6/2
cos2a=√2/2 2a∈[-2π,2π]
2a=±π/4或2a=±7π/4
解得
a=±π/8或a=±7π/8
（2）
y=√3cos2x+sin(2x-π/3)
=√3cos2x+sin2xcosπ/3-cos2xsinπ/3
=√3cos2x+1/2 sin2x-√3/2 cos2x
=1/2 sin2x+√3/2 cos2x
=sin2xcosπ/3+cos2x sinπ/3
=sin(2x+π/3)
令-π/2+2kπ≤2x+π/3≤π/2+2kπ得
-5π/6+2kπ≤2x≤π/6+2kπ
-5π/12+kπ≤x≤π/12+kπ
所以单调增区间为[-5π/12+kπ,π/12+kπ] （k∈Z）