Example 2 Let the set $A=\{1,2, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies $17 \mid(a+b)$, then $B$ is said to have property $P$.
(1) Find the number of binary subsets of $A$ that have property $P$;
(2) The number of a set of binary subsets of $A$, which are pairwise disjoint and have property $P$, is how many?
Analysis According to the problem, we classify the set using the remainder of division by 17.
Solution (1) Divide $1,2, \cdots, 366$ into 17 categories based on the remainder when divided by 17: $[0],[1], \cdots,[16]$ (where [i] represents the category of numbers that leave a remainder of $i$ when divided by 17). Since $366=17 \times 21+9$, the categories $[1], [2], \cdots, [9]$ each contain 22 numbers; the categories $[10],[11], \cdots,[16]$ and $[0]$ each contain 21 numbers.
(1) When $a, b \in[0]$, the number of subsets with property $P$ is $C_{21}^{2}=210$;
(2) When $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{21}^{1}=462$;
(3) When $a \in[8], b \in[9]$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{22}^{1}=484$.
Therefore, the number of subsets of $A$ with property $P$ is $210+462 \times 7+484=3928$.
(2) To ensure that the binary subsets do not intersect, when $a, b \in[0]$, 10 subsets can be paired; when $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, 21 subsets can be paired for each; when $a \in[8], b \in[9]$, 22 subsets can be paired.
Thus, the number of pairwise disjoint subsets with property $P$ is $10+21 \times 7+22=179$.