$2 \cdot 38$ Let set $A=\{1,2,3, \cdots, 366\}$. If a binary subset $B$ $=\{a, b\}$ of $A$ satisfies 17 । $(a+b)$, then $B$ is said to have property $P$.
(1) Find the number of binary subsets of $A$ that have property $P$;
(2) Find the number of pairwise disjoint binary subsets of $A$ that have property $P$.
[Solution] (1) $17 \mid (a+b)$ if and only if $a+b \equiv 0(\bmod 17)$, i.e., $a \equiv k(\bmod 17)$ and $b \equiv 17-k(\bmod 17)$ for $k=0,1,2, \cdots, 16$.

Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: $[0],[1], \cdots,[16]$. Since $366=17 \times 21+9$, the class $[0]$ contains 21 numbers, the classes $[1],[2], \cdots,[9]$ each contain 22 numbers, and the classes $[10],[11], \cdots,[16]$ each contain 21 numbers.
$17 \mid (a+b)$ if and only if $a, b \in[0]$ or $a \in[k], b \in[17-k], k=1,2, \cdots, 16$.
When $a, b \in[0]$, the number of subsets with property $P$ is $C_{21}^{2}=210$.
When $a \in[k], b \in[17-k]$, fixing $k \in\{1,2, \cdots, 7\}$, the number of subsets with property $P$ is $C_{22}^{1} \cdot C_{21}^{1}=462$.
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 total number of all two-element subsets of $A$ with property $P$ is
$$
210+462 \times 7+484=3928 \text {. }
$$
(2) To ensure that the two-element subsets are pairwise disjoint, we can take
10 subsets $\{a, b\}$ with $a \in[0], b \in[0]$;
21 subsets $\{a, b\}$ with $a \in[k], b \in[17-k]$, fixing $k \in\{1,2, \cdots, 7\}$; 22 subsets $\{a, b\}$ with $a \in[8], b \in[9]$.
Therefore, the total number of two-element subsets of $A$ with property $P$ that are pairwise disjoint is
$$
10+21 \times 7+22=179 \text { (subsets). }
$$