∵α、β∈(-π/2,0)
∴cosα>0,sinβ0,α+β∈(-π,0)
∵sinα=-√5/5,tanβ=-1/3
∴cosα=2/√5,sinβ=-1/√10,cosβ=3/√10
==>tanα=-1/2
则(1)∵tan(α+β)=(tanα+tanβ)/(1-tanα*tanβ)
=(-1/2-1/3)/(1-1/6)
=-1
∴α+β=-π/4；
(2)√2sin(π/4-α)+cos(π/4+β)=√2(sin(π/4)cosα-cos(π/4)sinα)+(cos(π/4)cosβ-sin(π/4)sinβ)
=cosα-sinα+(cosβ-sinβ)/√2
=2/√5+1/√5+(3/√10+1/√10)/√2
=√5.