64. It is known that $A \cap B=\{1 ; 2\}, A \cap C=\{2 ; 5\}, A \cup B=$ $=\{1 ; 2 ; 5 ; 6 ; 7 ; 9\}, B \cup C=\{1 ; 2 ; 3 ; 4 ; 5 ; 7 ; 8\}$. Find the sets $A, B$ and $C$.
$\triangle$ When solving, we will create a table. It is more convenient to fill it in by columns rather than by rows. For example, if the element 5 belongs to set $B$, we will place a plus at the intersection of the corresponding row and column of the table; otherwise, we will place a minus.

1) Let's start with the number 1. Since 1 belongs to the intersection of sets $A$ and $B$, then $1 \in A$ and $1 \in B$. But 1 does not belong to $C$, because if 1 belonged to $C$, then it would belong to the intersection of sets $A$ and $C$, which contradicts the condition.
2) Take the number 2. Similarly, $2 \in A, 2 \in B, 2 \in C$.
3) Take the number 3. Let's start here by noting that 3 does not belong to the union of sets $A$ and $B$. Therefore, this number does not belong to either of the sets $A$ and $B$. But since $3 \in B \cup C$, then 3 belongs to $C$.

Table 4

| Set Element $^{2}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $A$ | + | + | - | - | + | + | - | - | + |
| $B$ | + | + | - | - | - | - | + | - | - |
| $C$ | - | + | + | + | + | - | - | + | - |

4) For the number 4, we similarly get: $4 \notin A, 4 \notin B, 4 \in C$.
5) Consider the number $5: 5 \in A, 5 \in C, 5 \notin B$.
6) For the number 6: $6 \notin B, 6 \notin C, 6 \in A$.
7) The situation is more complicated with the number 7. Does this number belong to set $A$? Suppose $7 \in A$. Since $7 \in B \cup C$, then $7 \in B$ or $7 \in C$, and then respectively $7 \in A \cap B$ or $7 \in A \cap C$. However, both of these contradict the conditions of the problem. Therefore, we must conclude that the number 7 does not belong to $A$.

Since $7 \in A \cup B, 7 \notin A$, then 7 belongs to $B$.

As for $C$, 7 may belong to it, or it may not.

8) Take the number $8: 8 \notin A, 8 \notin B, 8 \in C$.
9) Finally, for the number 9 we get: $9 \notin B, 9 \notin C, 9 \in A$.

Answer: $A=\{1 ; 2 ; 5 ; 6 ; 9\}, B=\{1 ; 2 ; 7\}, C=\{2 ; 3 ; 4 ; 5 ; 7 ; 8\}$, or $A=\{1 ; 2 ; 5 ; 6 ; 9\}, B=\{1 ; 2 ; 7\}, C=\{2 ; 3 ; 4 ; 5 ; 8\}$.