Solution:

One spider cannot catch the fly. It suffices that the fly does not move if the spider is not on an adjacent field, that is, that it moves to the field diagonally opposite the spider's if it is.
We will prove that two spiders suffice in both parts of the problem - let us denote them by $P$ and $Q$, the fly by $M$, and the $x$- and $y$-coordinate of a point $A$ by $A_{x}$ and $A_{y}$.

a) Let us place the origin of the coordinate system in the lower left corner of the grid. At the start, moving along the $x$-axis, $P$ achieves in a finite number of moves that $P_{x}=M_{x}$. Analogously, $Q$ achieves that $Q_{y}=M_{y}$. Afterwards the spiders move in the following way: whenever the fly changes its $x$-coordinate, $P$ does the same so that $P_{x}=M_{x}$ remains true, otherwise it approaches the fly by one step along the $y$-axis; the movement of $Q$ is analogous. In this way the quantity $|P_{y}-M_{y}|+|Q_{x}-M_{x}|$ either decreases or stays the same, where it can stay the same for at most $2 \cdot 2010$ moves (when the fly is retreating). Therefore, after a finite number of moves at least one summand will become zero, i.e. the fly will be caught.

б) Disregarding the $z$-axis, based on part (a), one of the spiders, say $P$, can achieve that $P_{x}=M_{x}$ and $P_{y}=M_{y}$. Further, whenever the fly makes a step along the $z$-axis, the spider $P$ moves toward it, and otherwise it moves so as always to be exactly below the fly. It is clear that the fly can make a step along the $z$-axis, or stay in place, only finitely many times without being caught by the spider $P$. Therefore, starting from some moment, the fly must move exclusively within its $xy$-plane, without staying in place.
Now the spider $Q$ reaches, in a finite number of moves, the $xy$-plane in which the fly is moving. Also, by staying in place as needed, $Q$ achieves that $f=|Q_{x}-M_{x}|+|Q_{y}-M_{y}|$ is even. In each subsequent move, the spider $Q$ approaches the fly along the $x$-axis if $|Q_{x}-M_{x}|>|Q_{y}-M_{y}|$, and along the $y$-axis otherwise. After each move, the quantity $f$ does not increase and does not change its parity, and it can remain the same only finitely many times (when the fly is fleeing from the spider toward the edge of the grid). Therefore, at some moment $f=0$ will hold, and the fly will be caught.