8. For a plane $\alpha$, if the distances from all vertices of a polyhedron to the plane $\alpha$ are equal, then the plane $\alpha$ is called a "median plane" of the polyhedron.
(1) How many distinct median planes does a tetrahedron have?
(2) How many distinct median planes does a parallelepiped have?
(3) Given four non-coplanar points in three-dimensional space, how many distinct parallelepipeds can be formed using these four points as four of the vertices?
Please provide the answers to each of the above questions and explain the reasoning.
8. (1) Answer: 7.

Consider the tetrahedron as $ABCD$.
If all four vertices are on one side of the plane $\alpha$, then these four vertices must lie on a plane parallel to $\alpha$, which does not meet the condition.
We only consider the following two scenarios.
(i) Three vertices on one side of the plane $\alpha$ and one vertex on the other side.

Assume points $A, B, C$ are on one side of the plane $\alpha$ and point $D$ is on the other side. Then the plane determined by points $A, B, C$ must be parallel to plane $\alpha$.

Draw a perpendicular line $DD_1$ from point $D$ to plane $ABC$, with $D_1$ as the foot of the perpendicular. Then the median plane $\alpha$ must be the plane passing through the midpoint of $DD_1$ and perpendicular to $DD_1$ (exists and is unique), and this median plane is parallel to plane $ABC$.

Similarly, we can obtain the median planes parallel to the other three faces of the tetrahedron.
There are 4 such median planes.
(ii) Two vertices on each side of the plane $\alpha$. Assume points $A, B$ are on one side of the plane $\alpha$ and points $C, D$ are on the other side. Clearly, $AB \parallel \alpha, CD \parallel \alpha$.

It is easy to see that $AB$ and $CD$ are skew lines, and the median plane $\alpha$ must be the plane passing through the midpoint of their common perpendicular and parallel to them (exists and is unique). Since the 6 edges of the tetrahedron can be divided into 3 pairs of skew lines, there are 3 such median planes.

In summary, a tetrahedron has 7 distinct median planes.
(2) Answer: 3.
Since no five vertices of a parallelepiped are coplanar, the scenarios where the eight vertices of the parallelepiped are distributed on both sides of the plane $\alpha$ as $8-0, 7-1, 6-2, 5-3$ do not meet the condition.
We only consider the scenario where the eight vertices are four on each side of the plane $\alpha$. In this case, the 4 vertices on one side of the plane $\alpha$ must lie on a plane parallel to $\alpha$.
Consider the parallelepiped as $ABCD-A_1B_1C_1D_1$.
Since face $ABCD \parallel$ face $A_1B_1C_1D_1$, the plane passing through the midpoint of any common perpendicular of these two faces and parallel to these two faces is equidistant from these two faces. Thus, this plane is a median plane of the parallelepiped (exists and is unique). Similarly, we can obtain the other two median planes.
(3) Answer: 29.
From (2), we know that each parallelepiped has three median planes, which are parallel to the three pairs of opposite faces of the parallelepiped. For each parallelepiped, its three median planes are uniquely determined and intersect at one point (center $O$). Conversely, once three median planes intersecting at one point are given, the directions of the three pairs of opposite faces of the parallelepiped are completely determined. Combined with the given at least four vertices, the parallelepiped can be uniquely determined.
From (1), we know that for the given four points in space, there are seven distinct median planes. Choosing any three from these seven has $\mathrm{C}_7^3$ ways, but not all of these "median plane triplets" can determine a parallelepiped. In fact, three planes parallel to the same line cannot intersect at one point. In fact, they form the three sides of a triangular prism.
It is easy to see that among the seven median planes of the tetrahedron $ABCD$, exactly three are parallel to edge $AB$: the median plane parallel to the skew lines $AB$ and $CD$, the median plane parallel to plane $ABC$, and the median plane parallel to plane $ABD$. Thus, for each of the six edges of the tetrahedron $ABCD$, there are three median planes, and these six sets of median plane triplets cannot determine a parallelepiped. All other triplets of three median planes intersect at one point.
In summary, a total of $\mathrm{C}_7^3 - 6 = 29$ distinct parallelepipeds can be determined.