1.63 Given a set of points $E$ in coordinate space, consisting of all points with three coordinates being integers between 0 and 1982. Each point in $E$ is colored either red or blue, such that the number of red vertices on each rectangular parallelepiped with vertices in $E$ and edges parallel to the coordinate axes is divisible by 4. How many different coloring methods are there?
[Solution] First, we prove the following lemma.

Lemma: The coloring of points in set $E$ satisfies the requirements of the problem if and only if each rectangle with vertices in $E$ and edges parallel to the coordinate axes has an even number of red vertices.

Proof of the Lemma: If there is a rectangle $\pi_{0}$ with an odd number of red vertices, assume it has exactly 1 red vertex. Consider the two rectangular prisms that share $\pi_{0}$ as a common face, with the faces opposite to $\pi_{0}$ being $\pi_{1}$ and $\pi_{2}$. Since the coloring satisfies the problem's requirements, rectangles $\pi_{1}$ and $\pi_{2}$ each have 3 red vertices. Thus, the rectangular prism with $\pi_{1}$ and $\pi_{2}$ as a pair of opposite faces has 6 red vertices, which is a contradiction. This proves the necessity.

Assume each rectangle has an even number of red vertices. For any rectangular prism, if all its vertices are the same color, the number of red vertices is obviously divisible by 4. If the vertices are not all the same color, there must be an edge whose two endpoints are different colors. Since each rectangle has an even number of red vertices, the other 3 edges parallel to this edge in the rectangular prism also have two endpoints of different colors. Thus, the rectangular prism has exactly 4 red vertices. The lemma is proved.

Let $E_{1}$ be the set of $1 + 3 \times 1982$ points in the intersection of set $E$ with the 3 coordinate axes. For any coloring of $E_{1}$, there exists a unique coloring of $E$ that satisfies the problem's requirements and matches the given coloring on $E_{1}$. To this end, consider the function
$$
f(x, y, z)=\left\{\begin{array}{l}
0, \text { when point }(x, y, z) \text { is red, } \\
1, \text { when point }(x, y, z) \text { is blue. }
\end{array}\right.
$$

Determine the color of point $(x, y, z)$ based on the value of $f(x, y, z)$. Define the modulo 2 addition operation $a \oplus b$ as follows:
$$
a \oplus b=\left\{\begin{array}{l}
0, \text { when } a+b \text { is even. } \\
1, \text { when } a+b \text { is odd. }
\end{array}\right.
$$

For any $(x, y, z) \in E$, let
$$
f(x, y, z)=f(x, 0,0) \oplus f(0, y, 0) \oplus f(0,0, z) .
$$

It is easy to verify that this formula holds for points in $E_{1}$, and when the function values $f(x, y, z)$ are used to determine the coloring of points in $E - E_{1}$, each rectangle with edges parallel to the coordinate axes and vertices in $E$ has an even number of red points. In fact, when the vertices of the rectangle are $\left(x_{1}, y_{1}, z\right),\left(x_{2}, y_{1}, z\right),\left(x_{2}, y_{2}, z\right),\left(x_{1}, y_{2}, z\right)$, by definition we have
$$
\begin{array}{l}
f\left(x_{1}, y_{1}, z\right) \oplus f\left(x_{2}, y_{1}, z\right) \oplus f\left(x_{2}, y_{2}, z\right) \oplus f\left(x_{1}, y_{2}, z\right) \\
=f\left(x_{1}, 0,0\right) \oplus f\left(0, y_{1}, 0\right) \oplus f(0,0, z) \oplus f\left(x_{2}, 0,0\right) \oplus f\left(0, y_{1},\right. \\
\quad 0) \oplus f(0,0, z) \oplus f\left(x_{2}, 0,0\right) \oplus f\left(0, y_{2}, 0\right) \oplus f(0,0, z) \oplus \\
\quad f\left(x_{1}, 0,0\right) \oplus f\left(0, y_{2}, 0\right) \oplus f(0,0, z) \\
=0,
\end{array}
$$

which means the rectangle has an even number of red points. This shows that the coloring determined by the values of $f(x, y, z)$ satisfies the problem's requirements.

On the other hand, for any coloring of $E$ that satisfies the problem's requirements and matches the given coloring on $E_{1}$, consider the function value $f_{1}(x, y, z)$ of point $(x, y, z)$. By the lemma, the rectangle with vertices $(x, y, 0),(x, 0,0),(0, y, 0),(0,0,0)$ has an even number of red vertices, so we have
$$
f_{1}(x, y, 0)=f(x, 0,0) \oplus f(0, y, 0) \oplus f(0,0,0) .
$$

Similarly, we have
$$
f_{1}(x, 0, z)=f(x, 0,0) \oplus f(0,0,0) \oplus f(0,0, z) .
$$

Thus, we also have
$$
\begin{aligned}
& f_{1}(x, y, z)=f_{1}(x, y, 0) \oplus f(x, 0,0) \oplus f_{1}(x, 0, z) \\
= & f(x, 0,0) \oplus f(0, y, 0) \oplus f(0,0,0) \oplus f(x, 0,0) \oplus f(x, 0,0) \oplus \\
& f(0,0,0) \oplus f(0,0, z) \\
= & f(x, 0,0) \oplus f(0, y, 0) \oplus f(0,0, z)=f(x, y, z) .
\end{aligned}
$$

This proves that the correspondence from the coloring of $E_{1}$ to the coloring of $E$ that satisfies the problem's requirements is a bijection. Therefore, the number of different colorings of $E$ that satisfy the problem's requirements is equal to the number of colorings of $E_{1}$, which is $2^{1+3 \times 1982}=2^{5947}$.