7. Let the edge length of a regular tetrahedron be $2 \sqrt{6}$, and let a sphere be constructed with the center $O$ of the tetrahedron as its center. The total length of the curves formed by the intersection of the sphere's surface with the four faces of the tetrahedron is $4 \pi$. Then the radius of the sphere $O$ is $\qquad$.
7. $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$.

Let the radius of sphere $O$ be $R$.
If a face of the regular tetrahedron intersects the sphere as shown in Figure 3, then the circumference of the small circle is $\pi$, and the radius of the small circle is $\frac{1}{2}$.

Also, the distance from the center of the sphere to the face of the tetrahedron is 1, so
$$
\begin{array}{l}
R=\sqrt{1^{2}+\left(\frac{1}{2}\right)^{2}} \\
=\frac{\sqrt{5}}{2} .
\end{array}
$$

If a face of the regular tetrahedron intersects the sphere as shown in Figure 4, let $D$ be the midpoint of $A C$.

According to the problem, the length of arc $\overparen{A B}$ is $\frac{\pi}{3}$.

Let the radius of the small circle $\odot O_{1}$ be $r$.
Then $\angle A O_{1} B=\frac{\pi}{3 r}$.
$$
\begin{array}{l}
\text { Also, } \angle B O_{1} C=\frac{2 \pi}{3}, O_{1} D=\sqrt{2}, \\
\angle A O_{1} D=\frac{1}{2}\left(\angle B O_{1} C-\angle A O_{1} B\right) \\
=\frac{\pi}{3}-\frac{\pi}{6 r},
\end{array}
$$

Thus, $\cos \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)=\frac{\sqrt{2}}{r}$.
Let $f(r)=\cos \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)-\frac{\sqrt{2}}{r}$.
Then $f^{\prime}(r)=-\frac{\pi}{6 r^{2}} \sin \left(\frac{\pi}{3}-\frac{\pi}{6 r}\right)+\frac{\sqrt{2}}{r^{2}}>0$.
Therefore, the function $f(r)$ is monotonically increasing on the interval $(0,+\infty)$, and has at most one zero.
Since $f(2)=0$, the equation (1) has a unique solution 2.
Thus, $R=\sqrt{r^{2}+1}=\sqrt{5}$.
In summary, the radius of sphere $O$ is $\frac{\sqrt{5}}{2}$ or $\sqrt{5}$.