25. Four pairwise non-intersecting spheres of equal radii are given, with centers not lying in the same plane. a) How many spheres exist that touch all four spheres simultaneously? b) How to construct these spheres?
25. a) The contact of spheres can be of two kinds: 1) external contact, when the spheres lie one outside the other and the distance between their centers is equal to the sum of the radii; 2) internal contact, when one sphere is inside the other and the distance between the centers is equal to the difference of the radii.

A sphere $S$, touching the given four spheres, can occupy the following positions relative to them: 1) all 4 spheres touch sphere $S$ externally; 2) all 4 spheres touch sphere $S$ internally; 3) 3 spheres touch externally, one internally (4 positions); 4) 3 spheres touch internally, one externally (4 positions); 5) 2 spheres touch internally, two externally (6 positions). In total, for the touching sphere, we get 16 different possibilities.

b) Let us indicate the construction in each case.

1) Let a sphere with center $O$ and radius $R$ touch externally four spheres with centers $O_{1}, O_{2}, O_{3}, O_{4}$ and one radius $r$. Suppose the radii $r$ of the four spheres are reduced by the same length and the radius $R$ is increased by the same length. Since the distances between the centers have not changed, the contact of the spheres is preserved. Reduce the radii of the spheres to zero, then the spheres will "degenerate" into points $O_{1}, O_{2}, O_{3}, O_{4}$, and the radius of the tangent sphere will become equal to $R+r$. The problem is reduced to constructing a sphere through 4 given points. By constructing this sphere and reducing its radius by the length $r$, we obtain the desired sphere, touching the given four spheres.
2) If the radius of the sphere passing through the points $O_{1}, O_{2}, O_{3}, O_{4}$ is increased by $r$, then we obtain a sphere internally tangent to the four spheres.
3) Similarly, "contract" the three spheres of external contact to their centers; then the fourth tangent sphere will have a radius of $2r$. The problem is reduced to constructing a sphere passing through the given points and tangent to the given sphere. For this, we draw a circle through 3 points and draw a perpendicular to its plane through the center. The center of the tangent sphere must lie on this perpendicular. Draw a plane through this line and the center of the fourth sphere. The problem is reduced to a planimetric one: through two given points (these are the points of intersection of the obtained plane with the circle passing through the centers of the three spheres) draw a circle tangent to the given circle (see problem 49 from $\S 2$).
4) In this case, the three spheres are again contracted to points, and the further construction remains the same as in the previous case.
5) In this position, two spheres are contracted to two points, and the radii of the other two spheres are doubled. The center of the tangent sphere lies in the plane of symmetry of the two points and in the plane of symmetry of the two spheres. Since the 4 centers of the given spheres do not lie in the same plane, the indicated planes of symmetry intersect along some line $l$. Any sphere with its center on the line $l$ and passing through one of the points $O_{1}$ or $O_{2}$, due to symmetry, passes through the other as well, and if it touches one of the spheres, it touches the other as well. We again come to the same problem: through two given points, draw a sphere tangent to the given sphere. For its solution, the same planimetric construction can be used.