f(x)=cos(x-3/4π)-sin(x-5/4π)
=cosxcos3π/4+sinxsin3π/4-(sinxcos5π/4-cosxsin5π/4)
=-√2/2cosx+√2/2sinx-(-√2/2sinx+√2/2cosx)
=√2sinx-√2cosx
=2sin(x-π/4)
（1）最小正周期T=2π/1=2π
令π/2+2kπ≤x-π/4≤3π/2+2kπ (k∈Z)
3π/4+2kπ≤x≤7π/4+2kπ
单调递减区间为[3π/4+2kπ,7π/4+2kπ ] (k∈Z).
（2）
f(θ)=√2sinθ-√2cosθ=8/5
sinθ-cosθ=4√2/5
两边平方得
1-2sinθcosθ=32/25
2sinθcosθ=-7/25
于是
（sin2θ-2sin²θ）/（1-tanθ）
=(2sinθcosθ-2sin²θ）/（1-sinθ/cosθ）
=2sinθ(cosθ-sinθ）/[(cosθ-sinθ)/cosθ]
=2sinθcosθ
=-7/25