Yes, the characteristic equation is [itex]r^2+ 4r+ 3= (r+ 3)(r+ 1)= 0[/itex] which has roots r= -1 and r= -3 so the general solution to the associated homogeneous equation is [itex]Ae^{-3x}+ Be^{-x}[/itex].
For [itex]2e^{2x}cos(x)[/itex] try [itex]e^{2x}(Ccos(x)+ Dsin(x)[/itex]. For ##3xe^{-4x}##, try [itex]e^{-4x}(Ex+ F)[/itex]. For 3, try G.