(1)由tan(α+π/4)=2得,tanα=tan[(α+π/4)-π/4]=(2-1)/(1+2×1)= 1/3,所以,(2sinα+cosα)/(3cosα-2sinα)=(2tanα+1)/(3-2tanα)=(4+1)/(3-4)= -5.(2)由cosα=-(12/13),α∈(π,3π/2),得sinα=-(5/13),tanα=5...