答：
f(x)=sinx+acosx经过点（-π/3,0）
代入得：
f(-π/3)=sin(-π/3)+acos(-π/3)=0
所以：-√3/2+a/2=0
解得：a=√3
f(x)=sinx+√3cosx
=2*[(1/2)sinx+(√3/2)cosx]
=2sin(x+π/3)
g(x)=f²(x)-2
=4sin²(x+π/3)-2
=2*[1-cos(2x+2π/3)]-2
=-2cos(2x+2π/3)
g(x)最小正周期T=2π/2=π
单调递增区间满足：
2kπ