II. (40 points) Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$. Find
$$
\frac{\left(1+\cos ^{2} \alpha\right)\left(1-\sin ^{4} \alpha \cdot \cos ^{4} \beta\right)\left(1-\sin ^{4} \alpha \cdot \sin ^{4} \beta\right)}{\sin ^{2} 2 \alpha \cdot \sin ^{2} 2 \beta}
$$

the minimum value.
$$
\begin{array}{l}
\text { II. Let } x=\sin ^{2} \alpha \cdot \cos ^{2} \beta, y=\sin ^{2} \alpha \cdot \sin ^{2} \beta, \\
z=\cos ^{2} \alpha\left(\alpha, \beta \in\left(0, \frac{\pi}{2}\right)\right) .
\end{array}
$$

Then $x+y+z=1$. Therefore,
$$
\begin{array}{l}
\frac{\left(1+\cos ^{2} \alpha\right)\left(1-\sin ^{4} \alpha \cdot \cos ^{4} \beta\right)\left(1-\sin ^{4} \alpha \cdot \sin ^{4} \beta\right)}{\sin ^{2} 2 \alpha \cdot \sin ^{2} 2 \beta} \\
=\frac{(1+z)\left(1-\dot{x}^{2}\right)\left(1-y^{2}\right)}{16 \cdot \frac{x y z}{1-z}} \\
=\frac{\left(1-x^{2}\right)\left(1-y^{2}\right)\left(1-z^{2}\right)}{16 x y z} \\
=\frac{1-\left(x^{2}+y^{2}+z^{2}\right)}{16 x y z}+\frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{16 x y z}-\frac{x y z}{16} \\
=\frac{x y+y z+z x}{8 x y z}+\frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{16 x y z}-\frac{x y z}{16} \\
=\frac{1}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{16 x y z}-\frac{x y z}{16} \\
\geqslant \frac{1}{8} \cdot \frac{9}{x+y+z}+\frac{x y^{2} z+z^{2} x+z x^{2} y}{16 x y z}-\frac{1}{16 \times 3^{3}} \\
=\frac{32}{27} .
\end{array}
$$

Equality holds if and only if $x=y=z=\frac{1}{3}$. Therefore, the minimum value sought is $\frac{32}{27}$.