$$
\begin{array}{l}
\text { 2. Let } M=\frac{2 \cos 34^{\circ}-\cos 22^{\circ}}{\cos 14^{\circ}} \text {, } \\
N=\sin 56^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ} \text {. } \\
\text { Then } \frac{M}{N}=
\end{array}
$$
2.8.
$$
\begin{array}{l}
M=\frac{2 \cos 34^{\circ}-\sin 68^{\circ}}{\cos 14^{\circ}}=\frac{2 \cos 34^{\circ}\left(1-\sin 34^{\circ}\right)}{\cos 14^{\circ}} \\
=\frac{2 \cos 34^{\circ}\left(1-\cos 56^{\circ}\right)}{\cos 14^{\circ}}=\frac{4 \cos 34^{\circ} \cdot \sin ^{2} 28^{\circ}}{\cos 14^{\circ}} \\
=8 \cos 34^{\circ} \cdot \sin 28^{\circ} \cdot \sin 14^{\circ}=8 N .
\end{array}
$$

Therefore, $\frac{M}{N}=8$.
Note: A more general conclusion of this problem is
$$
\frac{2 \sin 4 \alpha-\sin 8 \alpha}{\cos \alpha}=8 \sin \alpha \cdot \sin 2 \alpha \cdot \sin 4 \alpha \text{. }
$$