(1)
∵向量m=((√3)sin(2x)+2,cosx),向量n=(1,2cosx)
∴f(x)
=向量m•向量n
=((√3)sin(2x)+2,cosx)•向量n=(1,2cosx)
=(√3)sin(2x)+2+2(cosx)^2
=(√3)sin(2x)+cos(2x)+3
=2sin(2x+π/6)+3
则最小正周期T=2π/2=π.
令π/2+2kπ≤2x+π/6≤3π/2+2kπ,得：π/6+kπ≤x≤2π/3+kπ(k∈Z)
则单调减区间为[π/6+kπ,2π/3+kπ](k∈Z).
(2)
∵f(x)=2sin(2x+π/6)+3
∴f(A)=2sin(2A+π/6)+3
∵f(A)=4
∴2sin(2A+π/6)+3=4
∴2sin(2A+π/6)=1
∴sin(2A+π/6)=1/2
∴2A+π/6=π/6或2A+π/6=5π/6
∵在△ABC中,A≠0
∴2A+π/6≠π/6
∴2A+π/6=5π/6
∴2A=2π/3
∴A=π/3
∴sinA=(√3)2,cosA=1/2
∵b=1
∴S△ABC=(1/2)bcsinA=[(√3)/4]c
∵S△ABC=(√3)/2
∴[(√3)/4]c=(√3)/2
∴c=2
∴a=√(b^2+c^2-2bccosA)=√(1^2+2^2-2×1×2×1/2)=√3.
(最后一步是余弦定理)