①f(x)=a·b
=(1,sin2x)·(cos2x,1)
=sin2x+cos2x
=(√2)sin(2x+π/4)
最小正周期=2π/2=π
②f(a/2)=(3√2)/5
即
(√2)sin(a+π/4)=(3√2)/5
亦即
sin(a+π/4)=3/5＞0
又 a∈(π/2,π)
∴ a+π/4∈(π/2,π)
∴ cos(a+π/4)=-4/5
∴ 1-2sin²a=cos2a
=sin(2a+π/2)
=2sin(a+π/4)cos(a+π/4)
=-24/25
∴ sina=(7√2)/10