28. Inscribed a regular tetrahedron in a sphere of radius $R$ (describe the construction). Find the volume of this tetrahedron.
28. The center $O$ of a regular tetrahedron divides its height in the ratio $1:3$. The center of the circumscribed sphere coincides with the center of the tetrahedron. Therefore, to construct a tetrahedron inscribed in a sphere, we choose any diameter of the sphere and at a distance of $\frac{1}{3} R$ from the center, we draw a plane perpendicular to it. We inscribe an equilateral triangle $ABC$ in the circle of intersection of this plane with the sphere, and connect the vertices of the triangle with the far end $S$ of the chosen diameter. We will prove that the regular triangular pyramid $SABC$ is a regular tetrahedron. Let $O'$ be the center of the circle of intersection. For example, by calculating the power of the point $O'$ relative to the sphere, we get:

$$
\begin{aligned}
& \frac{2 R}{3} \cdot \frac{4 R}{3} = \left|O' A\right|^2 ; \quad \left|O' A\right| = \frac{2 R \sqrt{2}}{3} \\
& \text{ Hence } |AB| = \frac{2 R \sqrt{2}}{3} \cdot \sqrt{3} = \frac{2 R \sqrt{6}}{3}
\end{aligned}
$$

On the other hand, by the Pythagorean theorem:

$$
\begin{gathered}
|SA|^2 = \left|O' S\right|^2 + \left|O' A\right|^2 = \frac{16 R^2}{9} + \frac{8 R^2}{9} = \frac{24 R^2}{9} \\
|SA| = \frac{2 R \sqrt{6}}{3}
\end{gathered}
$$

Thus, the sides of the base are equal in length to the lateral edges, i.e., the tetrahedron $SABC$ is regular. The area of its base is:

$$
\frac{1}{2} |AB|^2 \sin 60^\circ = \frac{1}{2} \cdot \frac{24 R^2}{9} \cdot \frac{\sqrt{3}}{2} = \frac{2 R^2 \sqrt{3}}{3}
$$

and the volume is

$$
V = \frac{1}{3} \cdot \frac{4}{3} R \cdot \frac{2 R^2 \sqrt{3}}{3} = \frac{8 R^3 \sqrt{3}}{27}
$$