Let $M$ be the set of all positive integers less than 1000. The operation “o” on $M$ is defined as follows:
Let $a, b \in M$, if $a b \in M$, then $a \circ b=a b$. If $a b \notin M$, let $a b=1000 k+r$, where $k$ is a positive integer, $r$ is a non-negative integer, and $r<1000$. When $k+r \in M$, $a \circ b=k+r$; when $k+r \notin M$, let $k+r=1000+s, a \circ b=s+1$.

For example, $559 \times 297=166023$, so $599 \circ 297=166+23=189$. Again, $559 \times 983=549497$, $549+497=1046$, so $559 \circ 983=1+46=47$.
(1) Find $559 \circ 758$;
(2) Find $x \in M$ such that $559 \circ x=1$;
(3) Is the operation associative? That is, for any $a, b, c \in M$, is it always true that $a \circ(b \circ c)=$ $(a \circ b) \circ c$? If it holds, please provide a proof; if not, please provide a counterexample.
(Single Dun, provided)
(1) $559 \times 758=423722, 423+722=1145>1000$, so, $559 \circ 758=145+1=146$.
(2) First, prove: for $a, b \in M$,
$$
a \circ b \equiv a b \quad(\bmod 999) .
$$

When $a b \in M$, equation (1) is obviously true.
When $a b \notin M$, $a b=1000 k+r$. If $k+r \in M$, then $a \circ b=k+r \equiv 1000 k+r \equiv a b(\bmod 999)$; if $k+r \notin M$, then $a \circ b=s+1=k+r-999 \equiv k+r \equiv 1000 k+r \equiv a b(\bmod 999)$. In both cases, equation (1) holds.
Next, prove: for $a, b \in M, a \circ b<1000$.
For $a, b \in M, a \cdot b<1000000$, let $a b=1000 k+r$, then $k<1000$, and $0 \leqslant r \leqslant 999$, so $k+r<1999$. If $k+r<1000$, then $a \circ b=k+r<1000$; if $1000 \leqslant k+r<1999$, then $a \circ b=k+r-999<1000$.
From equation (1), we know that $559 \circ x=1(x \in M)$ is equivalent to $559 x \equiv 1(\bmod 999)(x \in M)$.
Since $(559,999)=1$, according to the properties of complete residue systems, such an $x$ exists and is unique.
Since $9 \mid 999$ and $37 \mid 999$, we have $1 \equiv 559 x \equiv x(\bmod 9)$, so let $x=9 t+1(t \in \mathbf{N})$. Also, $1 \equiv 559 x \equiv 4 x \equiv 36 t+4 \equiv 1(\bmod 37)$, so $t \equiv 3(\bmod 37)$.
Upon inspection, we find that $x=361$.
(3) From equation (1), we know
$$
a \circ(b \circ c) \equiv a \cdot(b \circ c) \equiv a b c \quad(\bmod 999) .
$$

Similarly,
$$
(a \circ b) \circ c \equiv(a \circ b) \cdot c \equiv a b c \quad(\bmod 999) .
$$

Therefore, it satisfies the associative law.