Let $ Z$ and $ R$ denote the sets of integers and real numbers, respectively.
Let $ f: Z \rightarrow R$ be a function satisfying:
(i) $ f(n) \ge 0$ for all $ n \in Z$
(ii) $ f(mn)\equal{}f(m)f(n)$ for all $ m,n \in Z$
(iii) $ f(m\plus{}n) \le max(f(m),f(n))$ for all $ m,n \in Z$

(a) Prove that $ f(n) \le 1$ for all $ n \in Z$
(b) Find a function $ f: Z \rightarrow R$ satisfying (i), (ii),(iii) and $ 0<f(2)<1$ and $ f(2007) \equal{} 1$
### Part (a)

1. **Initial Conditions and Simplifications:**
   - Given \( f: \mathbb{Z} \rightarrow \mathbb{R} \) with:
     - \( f(n) \ge 0 \) for all \( n \in \mathbb{Z} \)
     - \( f(mn) = f(m)f(n) \) for all \( m, n \in \mathbb{Z} \)
     - \( f(m+n) \le \max(f(m), f(n)) \) for all \( m, n \in \mathbb{Z} \)

2. **Case Analysis for \( f(1) \):**
   - Let \( m = 1 \). Then \( f(1 \cdot n) = f(1)f(n) \implies f(n) = f(1)f(n) \).
   - If \( f(1) = 0 \), then \( f(n) = 0 \) for all \( n \in \mathbb{Z} \), which satisfies all conditions.
   - If \( f(1) \neq 0 \), then \( f(1) = 1 \) because \( f(n) = f(1)f(n) \implies f(1) = 1 \).

3. **Symmetry and Non-negativity:**
   - Let \( m = n = -1 \). Then \( f((-1)(-1)) = f(-1)f(-1) \implies f(1) = f(-1)^2 \).
   - Since \( f(1) = 1 \), we have \( 1 = f(-1)^2 \implies f(-1) = 1 \) (since \( f \ge 0 \)).

4. **Even Function Property:**
   - For any \( n \in \mathbb{Z} \), let \( m = -n \). Then \( f(n)f(-n) = f(-n^2) = f(n^2) \).
   - Since \( f(n^2) = f(n)^2 \) and \( f(-n) = f(n) \), \( f \) is an even function.

5. **Bounding \( f(n) \):**
   - We need to show \( f(n) \le 1 \) for all \( n \in \mathbb{Z} \).
   - Using \( f(m+n) \le \max(f(m), f(n)) \):
     - Let \( m = n = 1 \). Then \( f(2) \le \max(f(1), f(1)) = 1 \).
     - By induction, assume \( f(k) \le 1 \) for \( k \le n \). Then \( f(n+1) \le \max(f(n), f(1)) = 1 \).

6. **Conclusion:**
   - Therefore, \( f(n) \le 1 \) for all \( n \in \mathbb{Z} \).

\(\blacksquare\)

### Part (b)

1. **Finding a Specific Function:**
   - We need to find \( f: \mathbb{Z} \rightarrow \mathbb{R} \) such that:
     - \( 0 < f(2) < 1 \)
     - \( f(2007) = 1 \)

2. **Constructing the Function:**
   - Consider \( f(n) = c^{v_2(n)} \), where \( v_2(n) \) is the 2-adic valuation of \( n \) (the highest power of 2 dividing \( n \)) and \( 0 < c < 1 \).
   - This function satisfies:
     - \( f(n) \ge 0 \) for all \( n \in \mathbb{Z} \).
     - \( f(mn) = c^{v_2(mn)} = c^{v_2(m) + v_2(n)} = c^{v_2(m)}c^{v_2(n)} = f(m)f(n) \).
     - \( f(m+n) \le \max(f(m), f(n)) \) because the 2-adic valuation of \( m+n \) is at most the maximum of the 2-adic valuations of \( m \) and \( n \).

3. **Verifying the Conditions:**
   - Choose \( c \) such that \( 0 < c < 1 \).
   - For \( f(2) = c \), we have \( 0 < c < 1 \).
   - For \( f(2007) = 1 \), note that \( 2007 \) is odd, so \( v_2(2007) = 0 \) and \( f(2007) = c^0 = 1 \).

4. **Conclusion:**
   - The function \( f(n) = c^{v_2(n)} \) with \( 0 < c < 1 \) satisfies all the given conditions.

\(\blacksquare\)

The final answer is \( \boxed{ f(n) = c^{v_2(n)} } \) with \( 0 < c < 1 \).