(50 points) For the ternary tuple $(A, B, C)$, the following transformations can be performed: (i) any permutation; (ii) transform $(A, B, C)$ into $(2B + 2C - A, B, C)$. The initial state is given as the ternary tuple $(-1, 0, 1)$.
(1) Is it possible to obtain the ternary tuple $(2008, 2009, 2010)$ or $(2009, 2010, 2011)$ after a finite number of transformations? Explain your reasoning;
(2) It is known that the ternary tuple $(1, 2024, x)$ can be obtained after a finite number of transformations. Find all possible values of $x$ and explain your reasoning.
(1) $A+B+C(\bmod 2)$ remains unchanged under all transformations;

Therefore, $1+0+(-1)=0$ is even, and it is impossible to get $2008 + 2009 + 2010$ which is odd;
(2) $A+B+C(\bmod 4)$ remains unchanged or becomes $-(A+B+C)$ under all transformations;

Therefore, $1+0+(-1)=0(\bmod 4)$, and it is impossible to get $2009 + 2010 + 2011 \equiv 2(\bmod 4) ;$
(3) Inductive construction:
$$
\begin{aligned}
(-1,0,1) \rightarrow(0,1,3) \rightarrow(1,3,8) \rightarrow(1,8,15) \rightarrow \\
(1,15,24) \rightarrow \cdots \rightarrow\left(0,44^{2}-1=1935,45^{2}-1=2024\right) \\
\rightarrow\left(0,45^{2}-1,46^{2}-1=2115\right)
\end{aligned}
$$
$x$ can take the values 1935 or 2115;
To show that these are all possible values of $x$, we need to construct a (global) invariant. By Vieta's formulas for quadratic equations, we get
$A^{2}+B^{2}+C^{2}-2 A B-2 B C-2 C A$ remains unchanged under the transformations in the problem.