1. Given $\tan \alpha=m, \frac{3 \sin \alpha+\sin 3 \alpha}{3 \cos \alpha+\cos 3 \alpha}=$
1. $\frac{m}{2}\left(m^{2}+3\right)$.

Since $3 \sin \alpha+\sin 3 \alpha=2 \sin \alpha+\sin \alpha+\sin 3 \alpha=2 \sin \alpha+2 \sin \alpha \cos \alpha=2 \sin \alpha\left(1+2 \cos ^{2} \alpha\right)=$ $2 \sin \alpha\left(\sin ^{2} \alpha+3 \cos ^{2} \alpha\right), 3 \cos \alpha+\cos 3 \alpha=2 \cos \alpha+2 \cos \alpha \cos 2 \alpha=2 \cos \alpha[1+\cos 2 \alpha]=4 \cos ^{3} \alpha$, therefore $\frac{3 \sin \alpha+\sin 3 \alpha}{3 \cos \alpha+\cos 3 \alpha}=\frac{1}{2} \tan \alpha\left[\tan ^{2} \alpha+3\right]=\frac{1}{2} m\left(m^{2}+3\right)$.