高中向量数学，跪求。

1. a⊥b,sinxcosx-1/2=0
(sin2x)/2=1/2,sin2x=1
a+b=(sinx+cosx,1/2)
|a+b|^2=(sinx+cosx)^2+1/4=1+sin2x+1/4=9/4
|a+b|=3/2

2. f(x)=a^2-a.b=(sinx)^2+1-sinxcosx+1/2=1/2-(cos2x)/2+1-(sin2x)/2+1/2=2-(sin2x+cos2x)/2=2-2^(1/2)sin(2x+π/4)/2

2+2^(-1/2)>=f(x)>=2-2^(-1/2)

1》a*b=0
1/2sin(2x)=1/2
｜a+b｜^2=（sinx+cosx)^2+1/4
=5/4+sin2x
=9/4
｜a+b｜=3/2
2》f(x)=(sinx,1)(sinx-cosx,3/2)
=(sinx)^2-sinxcosx+3/2
=1-cos^2x-1/2sin2x+3/2
=5/2-（cos2x+1)/2+sin2x/2
=2-(cos2x-sin2x)/2
cos2x-sin2x取值区间为[-√2，√2]
值域为[2-√2,2+√2]

解：(1) ∵a⊥b ∴a·b = 0
｜a+b｜^2 = a^2 + b^2 + 2a·b =Sin^2 x + 1 + cos^2 x - 1/4 =7/4
(2) f(x)=a·（a-b）= sin^2 x - cosx-3/2 = (1-cos^2 x) - cosx-3/2=-cos^2 x- cosx-1/2 = -(cosx+1/2)^2-1/4
∵cosx ∈(-1,1)
∴f(x)max=f(1/2)=-1/4
f(x)min=f(1)=-5/2
∴f(x)的值域为〔-5/2，-1/4〕