(1)函数f(x)=asinwx+bcoswx(w>o)
由图示：T/2=11π/12-5π/12=π/2==>T=π==>w=2,f(x)初相角为第一象限角
设cosθ=a/√(a^2+b^2)
∴f(x)=√(a^2+b^2)sin(2x+θ)
f(-π/12)=√(a^2+b^2)sin(-π/6+θ)=0==>-π/6+θ=0==>θ=π/6
∴f(x)=√(a^2+b^2)sin(2x+π/6)
f(0)=√(a^2+b^2)sin(θ)=1==>sin(θ)=1/√(a^2+b^2)
∴(a^2+b^2)=2
∴f(x)=2sin(2x+π/6)
(2)解析：g(x)=f(x-π/12)-f(x+π/12)= 2sin(2x)-2sin(2x+π/3)
=sin(2x)-√3cos2x=2 sin(2x-π/3)
单调增区间：2kπ-π/2