（1）a*b=4cos(π/4-θ)*sin2θ*cos(π/4-θ)+cos4θ
=2[1+cos（π/2-2θ）]sin2θ+cos4θ
=2sin2θ+2(sin2θ)^2+cos4θ
=2sin2θ+1=(4√2+3)/3,
∴sin2θ=(2√2）/3,
2θ∈（π/2,π）,
cos2θ=-1/3,
tan2θ=-2√2.
（2）[2cos^（θ/2）-sinθ-1]/[√2sin（θ+π/4）]
=[cosθ-sinθ]/[sinθ+cosθ]
=[(cosθ)^2-(sinθ)^2]/(1+sin2θ)
=cos2θ/(1+sin2θ)
=-3+2√2.