∵sin(π/6+α)=3/5,α∈(π/3,5π/6) ,∴π/6+α∈(π/2,π),∴cos(π/6+α)=√1-[sin(π/6+α)]^2=√(1-9/25)=-4/5,∴tan(π/6+α)=sin(π/6+α)/cos(π/6+α)=-3/4,tan(5π/12+α)=tan[(π/6+α)+π/4]=[tan(π/6...