The function $f(x,y)$, defined on the set of all non-negative integers, satisfies 
(i) $f(0,y)=y+1$
(ii) $f(x+1,0)=f(x,1)$
(iii) $f(x+1,y+1)=f(x,f(x+1,y))$
Find f(3,2005), f(4,2005)
To solve the problem, we need to find the values of \( f(3,2005) \) and \( f(4,2005) \) for the function \( f(x,y) \) defined by the given conditions. We will use induction and the properties of the function to derive these values.

1. **Base Case and Induction for \( f(1, y) \):**
   - Given: \( f(0, y) = y + 1 \)
   - We claim: \( f(1, y) = y + 2 \)
   - **Base Case:** \( f(1, 0) = f(0, 1) = 2 \)
   - **Inductive Step:** Assume \( f(1, k) = k + 2 \) for \( 0 \leq k \leq n \). We need to show \( f(1, n+1) = n + 3 \).
     \[
     f(1, n+1) = f(0, f(1, n)) = f(0, n + 2) = (n + 2) + 1 = n + 3
     \]
   - Thus, by induction, \( f(1, y) = y + 2 \).

2. **Base Case and Induction for \( f(2, y) \):**
   - We claim: \( f(2, y) = 2y + 3 \)
   - **Base Case:** \( f(2, 0) = f(1, 1) = 3 \)
   - **Inductive Step:** Assume \( f(2, k) = 2k + 3 \) for \( 0 \leq k \leq n \). We need to show \( f(2, n+1) = 2n + 5 \).
     \[
     f(2, n+1) = f(1, f(2, n)) = f(1, 2n + 3) = (2n + 3) + 2 = 2n + 5
     \]
   - Thus, by induction, \( f(2, y) = 2y + 3 \).

3. **Base Case and Induction for \( f(3, y) \):**
   - We claim: \( f(3, y) = 2^{y+3} - 3 \)
   - **Base Case:** \( f(3, 0) = f(2, 1) = 5 \)
     \[
     2^{0+3} - 3 = 2^3 - 3 = 8 - 3 = 5
     \]
   - **Inductive Step:** Assume \( f(3, k) = 2^{k+3} - 3 \) for \( 0 \leq k \leq n \). We need to show \( f(3, n+1) = 2^{n+4} - 3 \).
     \[
     f(3, n+1) = f(2, f(3, n)) = f(2, 2^{n+3} - 3) = 2(2^{n+3} - 3) + 3 = 2^{n+4} - 6 + 3 = 2^{n+4} - 3
     \]
   - Thus, by induction, \( f(3, y) = 2^{y+3} - 3 \).

4. **Base Case and Induction for \( f(4, y) \):**
   - We claim: \( f(4, y) = \underbrace{2^{2^{2^{\ldots}}}}_{y+3 \text{ times}} - 3 \)
   - **Base Case:** \( f(4, 0) = f(3, 1) = 13 \)
     \[
     2^{2^2} - 3 = 2^4 - 3 = 16 - 3 = 13
     \]
   - **Inductive Step:** Assume \( f(4, k) = \underbrace{2^{2^{2^{\ldots}}}}_{k+3 \text{ times}} - 3 \) for \( 0 \leq k \leq n \). We need to show \( f(4, n+1) = \underbrace{2^{2^{2^{\ldots}}}}_{n+4 \text{ times}} - 3 \).
     \[
     f(4, n+1) = f(3, f(4, n)) = f(3, \underbrace{2^{2^{2^{\ldots}}}}_{n+3 \text{ times}} - 3) = \underbrace{2^{2^{2^{\ldots}}}}_{n+4 \text{ times}} - 3
     \]
   - Thus, by induction, \( f(4, y) = \underbrace{2^{2^{2^{\ldots}}}}_{y+3 \text{ times}} - 3 \).

The final answer is \( f(4, 2005) = \boxed{\underbrace{2^{2^{2^{\ldots}}}}_{2008 \text{ times}} - 3} \)