7.59 Let
$S=\left\{A=\left(a_{1}, a_{2}, \cdots, a_{8}\right) \mid a_{i}=0\right.$ or $\left.1, i=1,2, \cdots, 8\right\}$.
For any two elements $A$ and $B$ in $S$, define
$$
d(A, B)=\sum_{i=1}^{8}\left|a_{i}-b_{i}\right|,
$$

and call it the distance between $A$ and $B$. How many elements can be selected from $S$ at most, such that the distance between any two of them is not less than 5?
[Solution 1] The following 4 8-tuple sequences
$(0,0,0,0,0,0,0,0),(0,0,0,1,1,1,1,1)$,
$(1,1,1,1,1,0,0,0),(1,1,1,0,0,1,1,1)$
satisfy the requirements of the problem, so the maximum number of elements sought is no less than 4.
On the other hand, if there are 5 sequences $A, B, C, D, E$ with a distance of no less than 5 between any two, then without loss of generality, assume all 8 terms of $A$ are 0, thus each of $B, C, D, E$ has at least 5 terms as 1.

If two of $B, C, D, E$ have a distance of no less than 6 from $A$, i.e., at least 6 of the 8 terms are 1, then the distance between the two is no more than 4, a contradiction.

If two of $B, C, D, E$ have a distance of 5 from $A$, then the distance between the two must be 6. Without loss of generality, let
$$
\begin{array}{l}
B=(1,1,1,1,1,0,0,0), \\
C=(1,1,0,0,0,1,1,1) .
\end{array}
$$

Since the last 6 terms of $B$ and $C$ are all different, any 8-tuple sequence's last 6 terms will differ from at least one of $B$ or $C$ by at most 3 terms. Therefore, the first two terms must be 0, i.e., the first two terms of $D$ and $E$ are 0. To have a distance of no less than 5 from $A$, the last 6 terms of $D$ and $E$ must have at least 5 ones. Thus, $d(D, E) \leqslant 2$, a contradiction.

In summary, the maximum number of sequences that can be selected such that the distance between any two is no less than 5 is 4.
[Solution 2] From the example at the beginning of Solution 1, we know that the maximum number of elements sought is no less than 4.
If there are 5 sequences such that the distance between any two is no less than 5, then by the pigeonhole principle, there must be 3 sequences with the same 8th term. Without loss of generality, assume the 8th term of $A, B, C$ is 0. By symmetry, we can assume $A=(0,0,0,0,0,0,0,0)$. Thus, the first 7 terms of $B$ and $C$ must have at least 5 ones. Therefore, $d(B, C) \leqslant 4$, a contradiction.
In summary, the maximum number of sequences that can be selected such that the distance between any two is no less than 5 is 4.
[Solution 3] From the example at the beginning of Solution 1, we know that the maximum number of elements sought is no less than 4.
If there are 5 sequences such that the distance between any two is no less than 5, then write them as a $5 \times 8$ array
$$
\begin{array}{l}
a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8} \\
b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}, b_{8} \\
c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7}, c_{8} \\
d_{1}, d_{2}, d_{3}, d_{4}, d_{5}, d_{6}, d_{7}, d_{8} \\
e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}, e_{7}, e_{8}
\end{array}
$$

In the table, each column of 5 numbers $a_{i}, b_{i}, c_{i}, d_{i}, e_{i}$ can form 10 pairs. Since each number is either 0 or 1, there are at most 6 different pairs among the 10 pairs, i.e., $(0,1)$ or $(1,0)$. Therefore, among the 8 columns, there are at most 48 different pairs.

On the other hand, since the distance between any two sequences is no less than 5, the number of different pairs formed by the corresponding terms of any two sequences is at least 5. Thus, the 5 sequences can form at least 50 different pairs, a contradiction.
In summary, the maximum number of sequences that can be selected such that the distance between any two is no less than 5 is 4.