设函数f(x)=asin(kx+π/3)和g(x)=btan(kx-π/3)(k>0),若它们的最小正周期之和为3π/2,

最小正周期之和为3π/2,
2π/k+π/k=3π/2
k=2

f(π/2)=asin(π+π/3)=-a√3/2
g(π/2)=btan(π-π/3)=-b√3
所以，a=2b

f(π/4)=asin(π/2+π/3)=a/2
g(π/4)=btan(π/2-π/3)=b√3/3
所以，a/2=-b+1

解方程组得：
a=1,b=1/2
所以，
f(x)=sin(2x+π/3),g(x)=1/2*tan(2x-π/3)

f(x)=asin(kx+π/3)和g(x)=btan(kx-π/3)的最小正周期均是
T=2π/k
则 2T=4π/k =3π/2
k=8/3

f(π/2)=g(π/2),
f(π/2)=asin(8/3*π/2+π/3)=asin(5π/3)=-asin(2π/3)=-√3a/2
g(π/2)=btan(8/3*π/2-π/3)=btan(π)=0

a=0
我现在怀疑你的题是不是写错了