已知函数f(x)=cos^2(x+л/12),,g(x)=1+0.5sin2x

(1).设x=b 是函数y=f(x)的一条对称轴,求g(b)的值
(2).求函数h(x)=f(x)+g(x)的单调递增区间`

(1)
x=b 是函数y=f(x)的一条对称轴
对于任意实数m,有f(b+m)=f(b-m)
cos^2(b+m+π/12)=cos^2(b-m+π/12)
cos(b+m+π/12)=±cos(b-m+π/12)

cos(b+m+π/12)+cos(b-m+π/12)=0
cos(b+π/12)cos(m)=0,cos(b+π/12)=0
b+π/12=kπ+π/2,b=kπ+5π/12,k∈Z
g(b)=1+0.5sin(2kπ+5π/6)=1+0.5*0.5=5/4

或
cos(b+m+π/12)-cos(b-m+π/12)=0
sin(b+π/12)sin(m)=0,sin(b+π/12)=0
b+π/12=kπ,b=kπ-π/12,k∈Z
g(b)=1+0.5sin(2kπ-π/6)=1-0.5*0.5=3/4

g(b)=5/4或3/4

(2)h(x)=f(x)+g(x)
h'(x)=-2cos(x+π/12)*sin(x+π/12)+cos(x)
=cos(2x)-sin(2x+π/6)
=cos(2x)-cos(π/3-2x)
=-2sin(π/6)sin(2x-π/6)
令h'(x)>=0
sin(2x-π/6)<0
(2k-1)π<=2x-π/6<=2kπ
kπ-5π/12<=x<=kπ+π/12,k∈Z

h(x)单调递增区间[kπ-5π/12,kπ+π/12],k∈Z