2. $\frac{1-\sin ^{6} \theta-\cos ^{6} \theta}{1-\sin ^{4} \theta-\cos ^{4} \theta}=$
2. $\frac{3}{2}$.

Notice,
$$
\begin{array}{l}
\sin ^{4} \theta+\cos ^{4} \theta \\
=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cdot \cos ^{2} \theta \\
=1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta, \\
\sin ^{6} \theta+\cos ^{6} \theta \\
=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{4} \theta+\cos ^{4} \theta-\sin ^{2} \theta \cdot \cos ^{2} \theta\right) \\
=1-3 \sin ^{2} \theta \cdot \cos ^{2} \theta .
\end{array}
$$

Therefore, $\frac{1-\sin ^{6} \theta-\cos ^{6} \theta}{1-\sin ^{4} \theta-\cos ^{4} \theta}=\frac{3}{2}$.