It is given that f (x) is defined on R satisfying `f (1)=1 and ` for `AA x in R,`
`f (x+5) ge f (x) +5 and f (x+1) le f (x) +1`. If `g (x) =f (x) +1-x,` then `g (2002)`= __________

Answer

To solve the problem, we need to analyze the function \( f(x) \) based on the given conditions and then use it to find \( g(2002) \).

### Step 1: Analyze the conditions for \( f(x) \)

We are given:
1. \( f(1) = 1 \)
2. For all \( x \in \mathbb{R} \):
- \( f(x + 5) \geq f(x) + 5 \)
- \( f(x + 1) \leq f(x) + 1 \)

### Step 2: Establish relationships using the conditions

From the second condition, we can replace \( x \) with \( x + 4 \):
- \( f(x + 5) \leq f(x + 4) + 1 \)

We can continue this process:
- Replacing \( x \) with \( x + 3 \):
- \( f(x + 4) \leq f(x + 3) + 1 \)
- Replacing \( x \) with \( x + 2 \):
- \( f(x + 3) \leq f(x + 2) + 1 \)
- Replacing \( x \) with \( x + 1 \):
- \( f(x + 2) \leq f(x + 1) + 1 \)

Combining these inequalities, we can see that:
- \( f(x + 5) \leq f(x + 1) + 4 \)

### Step 3: Combine the inequalities

From the first condition:
- \( f(x + 5) \geq f(x) + 5 \)

Now we have:
- \( f(x) + 5 \leq f(x + 5) \leq f(x + 1) + 4 \)

### Step 4: Establish that \( f(x + 1) \) must equal \( f(x) + 1 \)

From the inequalities, we can deduce that:
- \( f(x + 1) \) must equal \( f(x) + 1 \) because if it were less than \( f(x) + 1 \), it would contradict the first inequality.

### Step 5: Conclude the form of \( f(x) \)

From the established relationship \( f(x + 1) = f(x) + 1 \), we can conclude that:
- \( f(x) = x + c \) for some constant \( c \).

Using the condition \( f(1) = 1 \):
- \( 1 + c = 1 \) implies \( c = 0 \).

Thus, we have:
- \( f(x) = x \).

### Step 6: Define \( g(x) \)

Now, we can define \( g(x) \):
- \( g(x) = f(x) + 1 - x \)
- Substituting \( f(x) = x \):
- \( g(x) = x + 1 - x = 1 \).

### Step 7: Find \( g(2002) \)

Since \( g(x) = 1 \) for all \( x \):
- \( g(2002) = 1 \).

### Final Answer

Thus, the value of \( g(2002) \) is:
\[ \boxed{1} \]