a=(cosα,sinα),b=(√2-sinα,cosα)
a·b=(cosα,sinα)·(√2-sinα,cosα)
=√2cosα-sinαcosα+sinαcosα
=√2cosα
1
|a+b|^2=|a|^2+|b|^2+2a·b=1+2+sinα^2-2√2sinα+cosα^2+2√2cosα
=4-2√2sinα+2√2cosα
=4-2√2(sinα-cosα)
=4-4sin(α-π/4)=4+2√3
即：sin(α-π/4)=-√3/2
α∈(-π/2,π/2),则：α-π/4∈(-3π/4,π/4)
故：α-π/4=-π/3或-2π/3
即：α=-π/12或-5π/12
2
c=(√2,sinα),则：a-c=(cosα-√2,0)
则：(a-c)·b=(cosα-√2,0)·(√2-sinα,cosα)
=√2cosα-sinαcosα-2+√2sinα
=-sin(2α)/2+2sin(α+π/4)-2
=cos(2α+π/2)/2+2sin(α+π/4)-2
=(1-2sin(α+π/4)^2)/2+2sin(α+π/4)-2
=-sin(α+π/4)^2+2sin(α+π/4)-3/2
=-(sin(α+π/4)-1)^2-1/2
α∈(-π/2,π/2),则：α+π/4∈(-π/4,3π/4)
即：sin(α+π/4)∈(-√2/2,1]
故当sin(α+π/4)=1时,(a-c)·b取得最大值：-1/2