Let

$$f(x)=4^{\sin ^{2} x}+8^{\cos ^{2} x}=4^{\sin ^{2} x}+8^{1-\sin ^{2} x}=2^{2 \sin ^{2} x}+8 \cdot 2^{-3 \sin ^{2} x}$$

Since the product of the cube of $2^{2 \sin ^{2} x}$ and the square of $2^{-3 \sin ^{2} x}$ is 1, we can divide $2^{2 \sin ^{2} x}$ into three "thirds" and $8 \cdot 2^{-3 \sin ^{2} x}$ into two "halves," resulting in the five quantities:

$$\frac{2^{2 \sin ^{2} x}}{3}, \quad \frac{2^{2 \sin ^{2} x}}{3}, \quad \frac{2^{2 \sin ^{2} x}}{3}, \quad 4 \cdot 2^{-3 \sin ^{2} x} \quad \text { and } \quad 4 \cdot 2^{-3 \sin ^{2} x}$$

The geometric mean of these five numbers does not depend on $x$:

$$G=\sqrt[5]{\left(\frac{2^{2 \sin ^{2} x}}{3}\right)^{3} \cdot\left(4 \cdot 2^{-2 \sin ^{2} x}\right)^{2}}=\sqrt[5]{\frac{16}{27}}$$

The arithmetic mean of these five numbers is $\frac{f(x)}{5}$, and since all five numbers are positive, the arithmetic mean-geometric mean inequality holds:

$$\frac{f(x)}{5} \geqq \sqrt[5]{\frac{16}{27}}, \quad \text { that is } \quad f(x) \geqq 5 \cdot \sqrt[5]{\frac{16}{27}}$$

Equality holds precisely when all five numbers are equal, i.e., $\frac{2^{2 \sin ^{2} x}}{3}=4 \cdot 2^{-3 \sin ^{2} x}$. This is equivalent to

$$\sin x= \pm \sqrt{\frac{1}{5} \log _{2} 12}= \pm \sqrt{\frac{\lg 12}{5 \lg 2}} \approx \pm 0.847$$

In the first quadrant,

$$x^{*}=\arcsin \sqrt{\frac{1}{5} \log _{2} 12} \approx \arcsin 0.847 \approx 1.0099$$

is the solution, and from this, all solutions of (2) are

$$x_{1}=x^{*}+k \pi, \quad x_{2}=-x^{*}+k \pi,(\mathrm{k} \text { integer })$$

At these points, the function will be minimal, and the minimum value is $5 \cdot \sqrt[5]{\frac{16}{27}}$.