5. In the tetrahedron $ABCD$, $DA=DB=DC=1$, and $DA$, $DB$, $DC$ are pairwise perpendicular. On the surface of the tetrahedron, the points that are a distance of $\frac{2 \sqrt{3}}{3}$ from point $A$ form a curve. The length of this curve is $\qquad$
5. $\frac{\sqrt{3} \pi}{2}$.

On the faces $DAB$, $DAC$, and $ABC$, the points that are a distance of $\frac{2 \sqrt{3}}{3}$ from point $A$ form circular arcs with a radius of $\frac{2 \sqrt{3}}{3}$. Let the points where these arcs intersect the edges $AB$, $AC$, $CD$, and $DB$ be $E$, $F$, $G$, and $H$ respectively.
From $AG = \frac{2 \sqrt{3}}{3}$,
$\Rightarrow DG = \sqrt{\left(\frac{2 \sqrt{3}}{3}\right)^2 - 1} = \frac{\sqrt{3}}{3}$
$\Rightarrow \angle GAD = \frac{\pi}{6}$.
Thus, the central angle of the arc $\overparen{GF}$ is $\frac{\pi}{12}$.
Similarly, the central angle of the arc $\overparen{EH}$ is also $\frac{\pi}{12}$.
Since $\triangle ABC$ is an equilateral triangle, the central angle of the arc $\overparen{EF}$ is $\frac{\pi}{3}$.
For the moving point on the face $DBC$ that is a distance of $\frac{2 \sqrt{3}}{3}$ from point $A$, the distance from point $D$ is $\frac{\sqrt{3}}{3}$. Therefore, the curve formed by the moving point on the face $DBC$ is a circular arc with $D$ as the center, $\frac{\sqrt{3}}{3}$ as the radius, and a central angle of $\frac{\pi}{2}$.
Thus, the length of this curve is
$$
\begin{array}{l}
\quad\left(\frac{\pi}{12}+\frac{\pi}{12}+\frac{\pi}{3}\right) \times \frac{2 \sqrt{3}}{3}+\frac{\pi}{2} \times \frac{\sqrt{3}}{3}=\frac{\sqrt{3} \pi}{2} . \\
=\frac{1-4^n}{1-4}=\frac{4^n-1}{3} . \\
\text{Let } n=2020, \text{ then } T_{2020}=\frac{4^{2020}-1}{3} .
\end{array}
$$

Let $n=2020$, then $T_{2020}=\frac{4^{2020}-1}{3}$.