(a) The answer is $f(3, 2005) = 2^{2008} - 3$.
We first label the equations as follows:
$$f(0, y) = y + 1, \qquad (1)$$
$$f(x + 1, 0) = f(x, 1), \qquad (2)$$
$$f(x + 1, y + 1) = f(x, f(x + 1, y)). \qquad (3)$$
By putting $y = 1$ in (1), we get $f(0, 1) = 2$. By putting $x = 0$ in (2), we get $f(1, 0) = f(0, 1) = 2$.
Using (3) and (1), we find that
$$f(1, y) = f(0, f(1, y - 1)) = f(1, y - 1) + 1.$$
By induction, one easily obtains
$$f(1, y) = f(1, 0) + y = y + 2.$$
Similarly, we have
$$f(2, y) = f(1, f(2, y - 1)) = f(2, y - 1) + 2.$$
By induction, we obtain
$$f(2, y) = f(2, 0) + 2y = f(1, 1) + 2y = 2y + 3.$$
Next, we have
$$f(3, y) = f(2, f(3, y - 1)) = 2f(3, y - 1) + 3.$$
Adding
$$\begin{aligned} f(3, y) &= 2f(3, y - 1) + 3, \\ 2f(3, y - 1) &= 2^2 f(3, y - 2) + 2 \cdot 3, \\ 2^2 f(3, y - 2) &= 2^3 f(3, y - 3) + 2^2 \cdot 3, \\ & \vdots, \\ 2^{y-1} f(3, 1) &= 2^y f(3, 0) + 2^{y-1} \cdot 3, \end{aligned}$$
we obtain
$$\begin{aligned} f(3, y) &= 2^y f(3, 0) + 3(1 + 2 + 2^2 + \cdots + 2^{y-1}) \\ &= 2^y f(2, 1) + 3(2^y - 1) = 2^{y+3} - 3. \end{aligned}$$
In particular, we have $f(3, 2005) = 2^{2008} - 3$.

(b) The answer is $f(4, 2005) = g(2008) - 3$, where $g(n)$ is defined by $g(1) = 2$ and $g(n + 1) = 2^{g(n)}$ for any $n \in \mathbb{Z}^+$.
We prove by induction that $f(4, y) = g(y + 3) - 3$. Firstly, by (2), we have
$$f(4, 0) = f(3, 1) = 2^4 - 3 = 2^{2^2} - 3 = g(3) - 3.$$
This proves the base case.
Assume $f(4, y) = g(y + 3) - 3$ for some $y \in \mathbb{Z}^+$. Using (3), we find that
$$f(4, y + 1) = f(3, f(4, y)) = 2^{f(4,y)+3} - 3 = 2^{g(y+3)} - 3 = g(y + 4) - 3.$$
This proves the inductive step.
Therefore, we have $f(4, 2005) = g(2008) - 3$.