∵cosα=-√5/5,α∈(π,3π/2)
∴sinα=√[1-(cosα)^2]=-2√5/5
∴tanα=sinα/cosα=2
∴tan(α-β)=(tanα-tanβ)/[1+tanαtanβ]
=[2-(1/3)]/[1-2×(1/3)]
=(5/3)/(1/3)
=5
∵α∈(π,3π/2),β∈(0,π/2)
∴α-β∈(π/2,3π/2)
∴α-β=arctan5+π/2