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 the above questions and explain the reasoning.

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(1) A tetrahedron has 1 distinct median plane.
   - **Reasoning**: A tetrahedron has 4 vertices. For a plane to be a median plane, it must be equidistant from all 4 vertices. The only plane that can satisfy this condition is the plane that passes through the centroid of the tetrahedron and is perpendicular to the line segment connecting the centroid to any vertex. This plane is unique, so there is only 1 distinct median plane for a tetrahedron.

(2) A parallelepiped has 3 distinct median planes.
   - **Reasoning**: A parallelepiped has 8 vertices. For a plane to be a median plane, it must be equidistant from all 8 vertices. The median planes of a parallelepiped are the planes that bisect the parallelepiped into two equal halves along the three pairs of opposite faces. Each pair of opposite faces defines a unique median plane, so there are 3 distinct median planes for a parallelepiped.

(3) Given four non-coplanar points in three-dimensional space, 3 distinct parallelepipeds can be formed using these four points as four of the vertices.
   - **Reasoning**: To form a parallelepiped, we need to choose 4 vertices from the 8 vertices of the parallelepiped. Given 4 non-coplanar points, we can form a parallelepiped by choosing one of the points as a vertex and then choosing the other 3 points as the adjacent vertices. The number of ways to choose 4 points from 8 is given by the combination formula \( \binom{8}{4} \). However, since the points are non-coplanar, we need to consider the specific arrangement of the points. Each set of 4 points can form a unique parallelepiped, and there are 3 distinct ways to choose the 4 points to form a parallelepiped. Therefore, there are 3 distinct parallelepipeds that can be formed.
8. (1) Answer: 7.

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

Assume points $A, B, C$ are on one side of 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 plane $\alpha$, assume points $A, B$ are on one side of 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 plane $\alpha$ as $8-0, 7-1, 6-2, 5-3$ do not meet the conditions.

We only consider the scenario where the eight vertices are on both sides of plane $\alpha$ with four points on each side. In this case, the 4 points on one side of plane $\alpha$ must lie on a plane parallel to plane $\alpha$.

Consider the parallelepiped as $ABCD-A_1B_1C_1D_1$.

Since plane $ABCD \parallel$ plane $A_1B_1C_1D_1$, the plane passing through the midpoint of any common perpendicular of these two planes and parallel to these two planes is equidistant from these two planes. 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 of these planes has $\mathrm{C}_7^3$ ways, but not all of these "median plane triplets" can determine a parallelepiped. In fact, the three planes that can form a parallelepiped must intersect at the same point, so we must exclude those triplets that do not intersect at the same point.

It is easy to see that three planes parallel to the same line cannot intersect at the same point. In fact, they form the three sides of a triangular prism.

It is easy to see that among the seven median planes of 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 tetrahedron $ABCD$, there are three median planes, and these six groups of median plane triplets cannot determine a parallelepiped. All other triplets of three median planes intersect at the same point.
In summary, a total of $\mathrm{C}_7^3 - 6 = 29$ distinct parallelepipeds can be determined.