(1)∵α∈(0.π/2),COSα=3/5
∴sinα>0,则sinα=4/5
∴tanα=4/3
故tan(π/4+α)
=[tan(π/4)+tanα]/[1-tan(π/4)*tanα]
=(1+4/3)/(1-4/3)
=-7
(2)∵α∈(0.π/2),β∈9(π/2,π),
∴α+β∈(π/2,3π/2),则cos(α+β)
(1)∵α∈(0.π/2),COSα=3/5
∴sinα>0,则sinα=4/5
∴tanα=4/3
故tan(π/4+α)
=[tan(π/4)+tanα]/[1-tan(π/4)*tanα]
=(1+4/3)/(1-4/3)
=-7
(2)∵α∈(0.π/2),β∈9(π/2,π),
∴α+β∈(π/2,3π/2),则cos(α+β)