函数f(x)=sin(wx+fai)在它的某一个周期内单调减区间[5π/12,11π5/12] |fai|T=π==>w=2π/π=2
∴f(x)=sin(2x+φ)
f(5π/12)=sin(2*5π/12+φ)=1==>5π/6+φ=π/2==>φ=-π/3
∴f(x)=sin(2x-π/3)

(2)解析：将y=f(x)的图像先向右平移π/6个单位
g(x)=sin(2(x-π/6)-π/3)=sin(2x-2π/3)
再将图像上的所有点的横坐标变为原来的1/2倍,纵坐标不变
T=π==>T=π/2==>w=4
∴g(x)=sin(4x-2π/3)
2kπ-π/2