(1) (This question is worth 35 points) Let the 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 all binary subsets of $A$ that have property $P$;
(2) Find the number of all pairwise disjoint binary subsets of $A$ that have property $P$.
(1) $a+b$ is divisible by 17 if and only if $a+b \equiv 0 \pmod{17}$, i.e., $\square$
$$
\begin{array}{l} 
a \equiv k \pmod{17}, b \equiv 17-k \pmod{17}, \\
k=0,1,2, \cdots, 16 .
\end{array}
$$

Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: [0], [1], ..., [16]. Since $366=17 \times 21+9$, there are 21 numbers in [0], 22 numbers in [1], [2], ..., [9], and 21 numbers in [10], ..., [16].

Let $17 \mid a+b$, if and only if $a, b \in [0]$ or $a \in [k], b \in [17-k]$.
When $a, b \in [0]$, the number of subsets with property $P$ is $C_{21}^{2}=210$.
When $a \in [k], b \in [17-k], k=1,2, \cdots, 7$, the number of subsets with property $P$ is $C_{22}^{1} \times C_{21}^{1}=462$ each.

When $a \in [8], b \in [9]$, the number of subsets with property $P$ is $C_{22}^{1} \times C_{22}^{1}=484$.
Therefore, the total number of two-element subsets of $A$ with property $P$ is
$$
210+462 \times 7+484=3928 \text{ (sets). }
$$
(2) To ensure that the two-element subsets are pairwise disjoint, we can pair them as follows:
$a \in [0], b \in [0]$, there are 10 subsets;
$a \in [k], b \in [17-k], k=1,2, \cdots, 7$, there are 21 subsets;
$a \in [8], b \in [9]$, there are 22 subsets.
Thus, the total number of two-element subsets of $A$ with property $P$ that are pairwise disjoint is
$$
10+21 \times 7+22=179 \uparrow .
$$