解(1)：
∵ tan(π/4+α)=(tanπ/4+tanα)/(1-tanπ/4×tanα)
=(1+tanα)/(1-tanα)
=3
∴ 3(1-tanα)=1+tanα
3-3tanα=1+tanα
4tanα=2
∴ tanα=1/2
(2)：
∵ tan2α=2tanα/(1-tan²α)
=(2×1/2)/[1-(1/2)²]
=1/(3/4)
=4/3
∴ cos2α≠0
∴(2sinαcosα+3cos2α)/(5cos2α-3sin2α)
=(sin2α+3cos2α)/(5cos2α-3sin2α)
=[(sin2α+3cos2α)×1/cos2α]/[(5cos2α-3sin2α)×1/cos2α]
=[(sin2α/cos2α)+3]/[5-3(sin2α/cos2α)]
=(tan2α+3)/(5-3tan2α)
=[(4/3)+3]/[5-3×(4/3)]
=(13/3)/(5-4)
=13/3