16提

g(x）=2sin(x-π/12+π/4)-√3+√3
=2sin(x+π/6)
f(π/4)+g(π/6)=2sin(π/4-π/12)-√3+2sin(π/6+π/6)=1-√3+√3=1
x=B
g(x)=2sin(B+π/6)有最大值
所以B+π/6=π/2
B=π/3
2accosB=a^2+c^2-b^2
b^2=a^2+c^2-ac
∵a+c=4
∴a^2+c^2+2ac=16
b^2=a^2+c^2-ac=16-3ac
(a+c)/2>=√ac (均值不等式）
ac=16-12=4
b^2>=4
b>=2
又因为b是三角形的边
∴a+c>b
b范围：2