8.140 There are three piles of stones with the numbers being $19,8,9$, respectively. Now the following operation is performed: each time, 1 stone is taken from any two piles, and then these 2 stones are added to the other pile. Try to determine:
(1) Whether it is possible to have the three piles of stones be $2,12,22$ after several such operations.
(2) Whether it is possible to have the three piles of stones all be 12 after several such operations.
If the requirements can be met, complete it with the minimum number of operations. If not, explain the reason.
［Solution］(1) It can be achieved.
Because the pile with the least stones originally, to make it become 2, at least six operations are required, in fact, six times can succeed:
$(19,8,9) \longrightarrow(21,7,8) \longrightarrow(23,6,7) \longrightarrow(22,5,9) \longrightarrow(24,4$,
$8) \longrightarrow(23,3,10) \longrightarrow(22,2,12)$.
(2) It cannot be achieved.
Since after each operation, the number of stones in each pile either decreases by 1 or increases by 2, we can write it as
$$
(a, b, c) \longrightarrow(a-1, b-1, c+2) \text {. }
$$

If $a, b, c$ are not congruent modulo 3, let's assume
$$
a \equiv 0(\bmod 3), b \equiv 1(\bmod 3), c \equiv 2(\bmod 3) \text {, }
$$

then after the operation, the three numbers $a-1, b-1, c+2$ are still not congruent modulo 3, i.e., $a-1 \equiv 2(\bmod 3), b-1 \equiv 0(\bmod 3), c+2 \equiv 1(\bmod 3)$.
Thus, each operation does not change the relationship that the remainders of these three numbers when divided by 3 are all different.

Since $19,8,9$ have remainders of $1,2,0$ when divided by 3, and $12,12,12$ have remainders of 0 when divided by 3, no matter how the operations are performed, it is impossible to make the number of stones in the three piles all 12.