12.3.8 ** Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1. A sphere is constructed with vertex $A$ as the center and $\frac{2 \sqrt{3}}{3}$ as the radius. The length of the curve formed by the intersection of the sphere's surface with the cube's surface is equal to
As shown in the figure, the sphere intersects with all six faces of the cube, resulting in two types of intersection lines: one type is on the three faces containing vertex $A$, i.e., faces $A A_{1} B_{1} B$, $A B C D$, and $A A_{1} D_{1} D$; the other type is on the three faces not containing vertex $A$, i.e., faces $B B_{1} C_{1} C$, $C C_{1} D_{1} D$, and $A_{1} B_{1} C_{1} D_{1}$.

On face $A A_{1} B_{1} B$, the intersection line is $\overparen{E F}$ and lies on the great circle passing through the center of the sphere $A$. Since $A E=\frac{2 \sqrt{3}}{3}$ and $A A_{1}=1$, then $\angle A_{1} A E=\frac{\pi}{6}$. Similarly, $\angle B A F=\frac{\pi}{6}$, so $\angle E A F=\frac{\pi}{6}$. Therefore, the length of $\overparen{E F}$ is $\frac{2 \sqrt{3}}{3} \cdot \frac{\pi}{6}=\frac{\sqrt{3}}{9} \pi$. There are three such arcs.

On face $B B_{1} C_{1} C$, the intersection line is $\overparen{F G}$ and lies on the small circle formed by the intersection of the plane at a distance of 1 from the center of the sphere and the sphere. In this case, the center of the small circle is $B$, with a radius of $\frac{\sqrt{3}}{3}$, and $\angle F B G=\frac{\pi}{2}$. Therefore, the length of $F G$ is $\frac{\sqrt{3}}{3} \cdot \frac{\pi}{2}=\frac{\sqrt{3}}{6} \pi$. There are also three such arcs. Thus, the total length of the curves is $3 \times \frac{\sqrt{3} \pi}{9}+3 \times \frac{\sqrt{3} \pi}{6}=\frac{5 \sqrt{3} \pi}{6}$.