Define a finite set $A$ to be 'good' if it satisfies the following conditions:

[list][*][b](a)[/b] For every three disjoint element of $A,$ like $a,b,c$ we have $\gcd(a,b,c)=1;$

[*][b](b)[/b] For every two distinct $b,c\in A,$ there exists an $a\in A,$ distinct from $b,c$ such that $bc$ is divisible by $a.$[/list]

Find all good sets.
To find all good sets \( A \) that satisfy the given conditions, we need to analyze the properties of such sets. Let's break down the problem step by step.

1. **Condition (a):** For every three disjoint elements \( a, b, c \in A \), we have \(\gcd(a, b, c) = 1\).

2. **Condition (b):** For every two distinct \( b, c \in A \), there exists an \( a \in A \), distinct from \( b \) and \( c \), such that \( bc \) is divisible by \( a \).

We will consider different cases based on the size of the set \( A \).

### Case 1: \( |A| = 3 \)

Let \( A = \{a, b, c\} \). We need to check if the set satisfies both conditions.

- **Condition (a):** Since \( A \) has only three elements, we need \(\gcd(a, b, c) = 1\). This means \( a, b, \) and \( c \) must be pairwise relatively prime.

- **Condition (b):** For any two elements \( b \) and \( c \), there must be an element \( a \) such that \( a \) divides \( bc \). This implies that \( a \) must be a common divisor of \( b \) and \( c \).

Given these conditions, one possible set is \( \{a, b, ab\} \) where \( \gcd(a, b) = 1 \). This set satisfies both conditions:
- \(\gcd(a, b, ab) = 1\)
- \( ab \) is divisible by both \( a \) and \( b \).

Another possible set is \( \{pq, qr, rp\} \) where \( p, q, r \) are pairwise relatively prime integers. This set also satisfies both conditions:
- \(\gcd(pq, qr, rp) = 1\)
- \( pq \) is divisible by \( p \) and \( q \), \( qr \) is divisible by \( q \) and \( r \), and \( rp \) is divisible by \( r \) and \( p \).

### Case 2: \( |A| \geq 4 \)

Suppose \( |A| \geq 4 \). Consider the smallest element \( c \) of \( A \). Using the choice of \( c \), observe that for any \( a \) in \( A \), there's a \( b \) such that \( \gcd(a, b) > 1 \), where \( a, b \neq c \).

Let \( A = \{a_1, a_2, \dots, a_n\} \) and let \( a_1 = c \). Consider the chain \( ca_2 \to ca_3 \to \cdots \to ca_n \) of dividends and distinguish two cases.

#### Case 2.1: For some \( a_i, a_j \in A \) other than \( c \), there exists an \( a \in A \) (\( a \neq c, a_i, a_j \)) such that \( a \mid ca_i \) and \( a \mid ca_j \).

- Then \( a \mid c \cdot \gcd(a_i, a_j) \) and since \( c < a \), there's a prime \( p \mid a \) that \( p \mid \gcd(a_i, a_j) \), which is a contradiction.

#### Case 2.2: For all \( a_i \in A \), the number \( a_j \) dividing \( ca_i \) is different than that of the others.

- Let \( a_{i_2} \mid ca_2, \dots, a_{i_n} \mid ca_n \). Then there's a permutation \( (j_2, \dots, j_n) \) of \( (i_2, \dots, i_n) \) such that \( a_{i_2} \mid c^2 a_{j_2}, \dots, a_{i_n} \mid c^2 a_{j_n} \).

- If for all \( 2 \leq i \leq n \), any prime \( q \mid a_i \) divides \( c \), then since there are \( a_i, a_j \) with \( \gcd(a_i, a_j) > 1 \), we can find a prime \( q \mid a_i, a_j, c \), which is absurd.

Thus, no set \( A \) with \( |A| \geq 4 \) can satisfy both conditions.

### Conclusion

The only good sets are those of the form \( \{a, b, ab\} \) with \( \gcd(a, b) = 1 \) and \( \{pq, qr, rp\} \) where \( p, q, r \) are pairwise relatively prime integers.

The final answer is \( \boxed{ \{a, b, ab\} } \) with \( \gcd(a, b) = 1 \) and \( \{pq, qr, rp\} \) where \( p, q, r \) are pairwise relatively prime integers.