Let's write down the problem: around a sphere of radius $r$, we want to construct a complete cone and a frustum of a cone, such that the volume of each is twice the volume of the sphere. Calculate the height and the radius of the base of the complete cone, as well as the radii of the bases of the frustum and the top of the frustum, and finally determine the ratio of the tangents of the slant heights.
Let the radius of the base of the cone be $AD = x$, its height $CD = m$, and $AC$ touches the sphere at $E$. According to the problem,

$$
\frac{8 r^{3} \pi}{3} = \frac{x^{2} \pi m}{3}
$$

but

$$
x = \frac{m r}{CE} = \frac{m r}{\sqrt{m^{2} - 2 m r}}
$$

Substituting this into (1):

$$
m^{2} - 8 r m = -16 r^{2}
$$

from which

$$
m = 4 r
$$

and thus

$$
x = r \sqrt{2}
$$

If, furthermore, the radius of the base of the frustum is $z = AB$, the radius of the top is $DC = v$, and $AD$ touches the sphere at $E$, then according to the problem:

$$
\frac{8 r^{2} \pi}{3} = \frac{2 r \pi}{3} (z^{2} + vz + v^{2})
$$

or

$$
4 r^{2} = z^{2} + vz + v^{2}
$$

but

$$
(z - v)^{2} + 4 r^{2} = (v + z)^{2}
$$

from which

$$
vz = r^{2}
$$

Substituting this into (2):

$$
z^{2} + v^{2} = 3 r^{2}
$$

From (3) and (4) we get that

$$
z = \frac{r}{2} (\sqrt{5} + 1) \text{ and } \frac{r}{2} (\sqrt{5} - 1)
$$

If the angles of inclination of the slant heights are denoted by $\alpha$ and $\beta$, then for the cone

$$
\tan \alpha = \frac{m}{x} = \frac{4 r}{r \sqrt{2}} = 2 \sqrt{2}
$$

and for the frustum

$$
\tan \beta = \frac{2 r}{z - v} = \frac{2 r}{r} = 2
$$

so

$$
\tan \alpha : \tan \beta = \sqrt{2} : 1
$$

(Krisztián György.)

The problem was also solved by: Czank K., Frank J., Kerekes T., Kiss A., Lukhaub Gy., Lupsa Gy., Obláth R., Sasvári G., Weisz J.