(2) Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a, b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor of $a$ and $c$ is 1, and the greatest common divisor of $b$ and $c$ is also 1;
(3) For any two distinct elements $a, b$ in $S$, there exists an element $d$ in $S$ different from $a$ and $b$ such that the greatest common divisor of $a$ and $d$ is greater than 1, and the greatest common divisor of $b$ and $d$ is also greater than 1.
The maximum number is 72.
Express each positive integer $n$ not exceeding 100 as
$$
n=2^{a_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdot q
$$

where $q$ is a positive integer not divisible by $2,3,5,7,11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-negative integers.

We select the positive integers that satisfy the condition "exactly 1 or 2 of $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-zero" to form the set $S$. This includes 50 even numbers $2,4, \cdots, 98,100$, but excluding $2 \times 3 \times 5,2^{2} \times 3 \times 5,2 \times 3^{2} \times 5,2 \times 3 \times 7,2^{2} \times 3 \times 7,2 \times 5 \times 7$, $2 \times 3 \times 11$, a total of 7 numbers; 17 odd multiples of 3, $3 \times 1,3 \times 3, \cdots, 3 \times 33$; 7 numbers with the smallest prime factor 5, $5 \times 1,5 \times 5,5 \times 7,5 \times 11,5 \times 13,5 \times 17,5 \times$ 19; 4 numbers with the smallest prime factor 7, $7 \times 1,7 \times 7,7 \times 11,7 \times 13$; and the prime number 11. Thus, $S$ contains a total of $(50-7)+17+7+4+1=72$ numbers.
We now prove that the constructed $S$ satisfies the given conditions.
Condition (1) is obviously satisfied.
For condition (2), note that among the prime factors of $[a, b]$, at most 4 of $2,3,5,7,11$ appear. Let the missing number be $p$. Clearly, $p \in S$, and
$$
\begin{array}{l}
(p, a) \leqslant(p,[a, b])=1 \\
(p, b) \leqslant(p,[a, b])=1
\end{array}
$$

Thus, taking $c=p$ suffices.
For condition (3), when $(a, b)=1$, take the smallest prime factor $p$ of $a$ and the smallest prime factor $q$ of $b$. Clearly, $p \neq q$, and $p, q \in\{2,3,5,7,11\}$. Thus, $p q \in S$, and
$$
(p q, a) \geqslant p>1, (p q, b) \geqslant q>1
$$
The fact that $a$ and $b$ are coprime ensures that $p q$ is different from $a$ and $b$. Thus, taking $d=p q$ suffices.
When $(a, b)=e>1$, take $p$ as the smallest prime factor of $e$, and $q$ as the smallest prime such that $q \times [a, b]$ is minimized. Clearly, $p \neq q$, and $p, q \in\{2,3,5,7,11\}$. Thus, $p q \in S$, and
$$
\begin{array}{l}
(p q, a) \geqslant(p, a)=p>1 \\
(p q, b) \geqslant(p, b)=p>1
\end{array}
$$
The condition $q \times [a, b]$ ensures that $p q$ is different from $a$ and $b$. Thus, taking $d=p q$ suffices.
We now prove that the number of elements in any set $S$ satisfying the given conditions does not exceed 72.

Clearly, $1 \notin S$. For any two primes $p, q$ greater than 10, since the smallest number not coprime with both $p$ and $q$ is $p q$, which is greater than 100, by condition (3), at most one of the 21 primes between 10 and 100, $11,13, \cdots, 89,97$, can appear in $S$. Let the set of the remaining 78 natural numbers not exceeding 100, excluding 1 and these 21 primes, be $T$. We claim that at least 7 numbers in $T$ are not in $S$, thus $S$ can have at most $78-7+1=72$ elements.
(1) If there is a prime $p$ greater than 10 in $S$, then the smallest prime factor of all numbers in $S$ can only be $2,3,5,7$, and $p$. Using condition (2), we can derive the following conclusions:
i If $7 p \in S$, since $2 \times 3 \times 5,2^{2} \times 3 \times 5,2 \times 3^{2} \times 5$ include all the smallest prime factors, by condition (2), $2 \times 3 \times 5,2^{2} \times 3 \times 5,2 \times 3^{2} \times 5 \notin S$; if $7 p \notin S$, note that $2 \times 7 p>100$, and $p \in S$, so by condition (3), $7 \times 1,7 \times 7,7 \times 11,7 \times 13 \notin S$.
ii If $5 p \in S$, then $2 \times 3 \times 7,2^{2} \times 3 \times 7 \notin S$; if $5 p \notin S$, then $5 \times 1,5 \times 5 \notin S$.
iii $2 \times 5 \times 7$ and $3 p$ cannot both be in $S$.
iv $2 \times 3 p$ and $5 \times 7$ cannot both be in $S$.
v If $5 p, 7 p \notin S$, then $5 \times 7 \notin S$.
When $p=11$ or 13, by i, ii, iii, iv, we can respectively conclude that at least $3,2,1,1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p=17$ or 19, by i, ii, iii, we can respectively conclude that at least $4,2,1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p>20$, by i, ii, iii, we can respectively conclude that at least $4,2,1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p>20$, by i, ii, iii, we can respectively conclude that at least $4,2,1$ numbers in $T$ are not in $S$, totaling 7 numbers.
(2) If no prime greater than 10 is in $S$, then the smallest prime factor of numbers in $S$ can only be $2,3,5,7$. Thus, the following 7 pairs of numbers cannot both be in $S$: $(3,2 \times 5 \times 7),(5,2 \times 3 \times 7),(7,2 \times 3 \times 5),(2 \times 3,5 \times 7)$,
$(2 \times 5,3 \times 7),(2 \times 7,3 \times 5),\left(2^{2} \times 7,3^{2} \times 5\right)$.
Thus, at least 7 numbers in $T$ are not in $S$.
In summary, the answer to this problem is 72.