(1)由題設(shè)知：
f(x)＝cos²(x＋π/12)＝1/2[1＋cos(2x＋π/6)]
∵x＝x0是函數(shù)y＝f(x)圖像的一條對稱軸
∴2x0＝kπ－π/6(k∈Z)
∴g(x0)＝1＋1/2sin2x0＝1＋1/2sin(kπ－π/6)
當k為偶數(shù)時,g(x0)＝1＋1/2sin(－π/6)＝1－1/4＝3/4
當k為奇數(shù)時,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
當2kπ－π/2≤2x＋π/3≤2kπ＋π/2
即：kπ－5π/12≤x≤kπ＋π/12(k∈Z)時,
函數(shù)h(x)＝1/2sin(2x＋π/3)＋3/2是增函數(shù)
當2kπ＋π/2≤2x＋π/3≤2kπ＋3π/2
即：kπ＋π/12≤x≤kπ＋7π/12(k∈Z)時,
函數(shù)h(x)＝1/2sin(2x＋π/3)＋3/2是減函數(shù)
故函數(shù)h(x)＝1/2sin(2x＋π/3)＋3/2的單調(diào)遞增區(qū)間是：[kπ－5π/12,kπ＋π/12] (k∈Z)
函數(shù)h(x)＝1/2sin(2x＋π/3)＋3/2的單調(diào)遞減區(qū)間是：[kπ＋π/12,kπ＋7π/12] (k∈Z)