(1)由题设知:
f(x)=cos²(x+π/12)=1/2[1+cos(2x+π/6)]
∵x=x0是函数y=f(x)图像的一条对称轴
∴2x0=kπ-π/6(k∈Z)
∴g(x0)=1+1/2sin2x0=1+1/2sin(kπ-π/6)
当k为偶数时,g(x0)=1+1/2sin(-π/6)=1-1/4=3/4
当k为奇数时,g(x0)=1+1/2sinπ/6=1+1/4=5/4
(2)
h(x)=f(x)+g(x)=1/2[1+cos(2x+π/6)]+1+1/2sin2x
=1/2[cos(2x+π/6)+sin2x]+3/2=1/2(√3/2 cos2x+1/2 sin2x)+3/2
=1/2sin(2x+π/3)+3/2
当2kπ-π/2≤2x+π/3≤2kπ+π/2
即:kπ-5π/12≤x≤kπ+π/12(k∈Z)时,
函数h(x)=1/2sin(2x+π/3)+3/2是增函数
当2kπ+π/2≤2x+π/3≤2kπ+3π/2
即:kπ+π/12≤x≤kπ+7π/12(k∈Z)时,
函数h(x)=1/2sin(2x+π/3)+3/2是减函数
故函数h(x)=1/2sin(2x+π/3)+3/2的单调递增区间是:[kπ-5π/12,kπ+π/12] (k∈Z)
函数h(x)=1/2sin(2x+π/3)+3/2的单调递减区间是:[kπ+π/12,kπ+7π/12] (k∈Z)