因为 $\cos36^{\circ}\sin18^{\circ}=\cos36^{\circ}\cos72^{\circ}=\dfrac{\sin72^{\circ}\cos72^{\circ}}{2\sin36^{\circ}}=\dfrac{\sin144^{\circ}}{4\sin36^{\circ}}=\dfrac{1}{4}$，$\cos36^{\circ}-\sin18^{\circ}=\sin54^{\circ}-\sin18^{\circ}=\sin(36^{\circ}+18^{\circ})-\sin(36^{\circ}-18^{\circ})=2\cos36^{\circ}\sin18^{\circ}=\dfrac{1}{2}$，令 $x=\cos36^{\circ}$，得 $\sin18^{\circ}=\dfrac{1}{4x}$，所以 $x-\dfrac{1}{4x}=\frac{1}{2}$，解得 $x=\dfrac{\sqrt{5}+1}{4}$，即 $\cos36^{\circ}=\frac{\sqrt{5}+1}{4}$．于是，$\dfrac{\sin^242^{\circ}\cos^212^{\circ}}{3\cos36^{\circ}+1}=\dfrac{(\sin42^{\circ}\cos12^{\circ})^2}{3\cos36^{\circ}+1}=\dfrac{\frac{1}{4}[\sin(42^{\circ}+12^{\circ})+\sin(42^{\circ}-12^{\circ})]^2}{3\cos36^{\circ}+1}$ $=\dfrac{\frac{1}{4}(\sin54^{\circ}+\sin30^{\circ})^2}{3\cos36^{\circ}+1}=\dfrac{\frac{1}{4}(\cos36^{\circ}+\frac{1}{2})^2}{3\cos36^{\circ}+1}=\dfrac{\frac{1}{4}(\frac{\sqrt{5}+1}{4}+\frac{1}{2})^2}{3\times\frac{\sqrt{5}+1}{4}+1}=\dfrac{\frac{1}{32}(7+3\sqrt{5})}{\frac{1}{4}(7+3\sqrt{5})}=\dfrac{1}{8}$．