Four. (50 points) A fly and $k$ spiders are placed at some intersections of a $2012 \times 2012$ grid. An operation consists of the following steps: first, the fly moves to an adjacent intersection or stays in place, then each spider moves to an adjacent intersection or stays in place (multiple spiders can occupy the same intersection). Assume that each spider and the fly always know the positions of the other spiders and the fly.
(1) Find the smallest positive integer $k$ such that the spiders can catch the fly in a finite number of operations, regardless of their initial positions;
(2) In a $2012 \times 2012 \times 2012$ three-dimensional grid, what is the conclusion for (1)?

[Note] In the problem, adjacent means that one intersection has only one coordinate different from another intersection's corresponding coordinate, and the difference is 1; catching means the spider and the fly are at the same intersection.
The minimum value of $k$ is 2.
(1) First, prove that a single spider cannot catch the fly.

Establish a Cartesian coordinate system, then the range of the spider and fly's movement is
$$
\{(x, y) \mid 0 \leqslant x, y \leqslant 2012, x, y \in \mathbf{N}\}.
$$

For each point in the above set, there are at least two points in the set that are adjacent to it. Therefore, if the spider is not adjacent to the fly after a move, the fly can remain still; if the spider is adjacent to the fly after a move, the fly can move to another adjacent point in the set. This way, the spider cannot catch the fly.

Second, if there are two spiders $S_{1}$ and $S_{2}$, let the distance between $S_{i}$ and the fly $F$ after $n$ moves be $a_{i, n} (i=1,2)$, where the distance between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is defined as
$$
d=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|.
$$

First, it can be achieved through several operations that
$$
S_{1}(0,0), S_{2}(2012,2012).
$$

Next, we prove:
(i) After each move, $S_{1}$ and $S_{2}$ can be appropriately moved so that the fly $F$ is inside or on the boundary of the rectangle with $S_{1}$ and $S_{2}$ as the diagonal (the line connecting $S_{1}$ and $S_{2}$ is not parallel to the coordinate axes);
(ii) Keep $a_{1, n}+a_{2, n}$ non-increasing, and it will strictly decrease after several moves.

In fact, let the unit vectors in the positive directions of the $x$ and $y$ axes be $\boldsymbol{i}$ and $\boldsymbol{j}$. The strategy is as follows:

If the fly's coordinate value is not inside or on the boundary of the rectangle with $S_{1}$ and $S_{2}$ as the diagonal after a move, then $S_{1}$ and $S_{2}$ move in that direction towards the grid where $\boldsymbol{F}$ is located; otherwise, $S_{1}$ can move $\boldsymbol{i}$ or $\boldsymbol{j}$, and $S_{2}$ can move $\boldsymbol{i}$ or $-\boldsymbol{j}$, appropriately chosen to ensure that the line connecting $S_{1}$ and $S_{2}$ is not parallel to the coordinate axes, and the fly's position is inside or on the boundary of the rectangle. By induction, (i) is easily proven.
Next, we prove (ii).
In fact, if the fly is strictly inside the rectangle with $S_{1}$ and $S_{2}$ as the diagonal before the $n$-th operation, then $a_{1, n}+a_{2, n}$ must decrease after that round; otherwise, the fly must be on one of the rectangle's sides and must move outside the rectangle, and by the rule, $a_{1, n}+a_{2, n}$ does not increase, but the grid is bounded, so after at most 4024 steps, the fly can no longer move outside the rectangle. At this point, $a_{1, n}+a_{2, n}$ strictly decreases.

Since the initial value of $a_{1,0}+a_{2,0}$ is 4024, it cannot decrease indefinitely, so after a finite number of operations, the spider can catch the fly.
(2) The conclusion remains unchanged, and the minimum value of $k$ is still 2.
Establish a spatial Cartesian coordinate system, then the range of $S_{i}$ and $F$ is $\{(x, y, z) \mid 0 \leqslant x, y, z \leqslant 2012, x, y, z \in \mathbf{N}\}$.
When $k=1$, the fly can adopt a similar strategy to ensure it is not caught;

When $k=2$, let $S_{1}$ and $S_{2}$ move to $(0,0,0)$ and $(2012,2012,2012)$ through several operations, and each operation keeps the fly inside or on the boundary (face, edge, vertex) of the cube with $S_{1}$ and $S_{2}$ as the body diagonal, and the line connecting $S_{1}$ and $S_{2}$ is not parallel to the planes $x O y$, $y O z$, and $z O x$. It can be completely similarly proven that after each operation, $a_{1, n}+a_{2, n}$ does not increase, and after at most $3 \times 2012$ steps, it will strictly decrease.

Since the initial value of $a_{1,0}+a_{2,0}$ is 6036, after a finite number of operations, the spider will catch the fly.