For any number $n\in\mathbb{N},n\ge 2$, denote by $P(n)$ the number of pairs $(a,b)$ whose elements are of positive integers such that 
\[\frac{n}{a}\in (0,1),\quad \frac{a}{b}\in (1,2)\quad \text{and}\quad \frac{b}{n}\in (2,3). \]
$a)$ Calculate $P(3)$.
$b)$ Find $n$ such that $P(n)=2002$.
Let's break down the problem and solution step-by-step.

### Part (a): Calculate \( P(3) \)

1. **Identify the constraints:**
   - \(\frac{n}{a} \in (0, 1)\) implies \(a > n\).
   - \(\frac{a}{b} \in (1, 2)\) implies \(b < a < 2b\).
   - \(\frac{b}{n} \in (2, 3)\) implies \(2n < b < 3n\).

2. **Substitute \( n = 3 \):**
   - From \(\frac{n}{a} \in (0, 1)\), we get \(a > 3\).
   - From \(\frac{a}{b} \in (1, 2)\), we get \(b < a < 2b\).
   - From \(\frac{b}{n} \in (2, 3)\), we get \(6 < b < 9\).

3. **Determine valid pairs \((a, b)\):**
   - For \(b = 7\):
     - \(a\) must satisfy \(7 < a < 14\). Possible values: \(a = 8, 9, 10, 11, 12, 13\).
   - For \(b = 8\):
     - \(a\) must satisfy \(8 < a < 16\). Possible values: \(a = 9, 10, 11, 12, 13, 14, 15\).

4. **Count the pairs:**
   - For \(b = 7\), there are 6 possible values for \(a\).
   - For \(b = 8\), there are 7 possible values for \(a\).

Thus, \(P(3) = 6 + 7 = 13\).

### Part (b): Find \( n \) such that \( P(n) = 2002 \)

1. **General formula for \( P(n) \):**
   - From the given solution, we have:
     \[
     P(n) = \sum_{b=2n+1}^{3n-1}(b-1) = \sum_{k=2n}^{3n-2}k = \frac{(3n-2)(3n-1) - 2n(2n-1)}{2}
     \]
   - Simplify the sum:
     \[
     P(n) = \frac{(3n-2)(3n-1) - 2n(2n-1)}{2} = \frac{9n^2 - 5n + 2 - 4n^2 + 2n}{2} = \frac{5n^2 - 3n + 2}{2}
     \]
   - Therefore:
     \[
     P(n) = \frac{5n^2 - 3n + 2}{2}
     \]

2. **Set \( P(n) = 2002 \):**
   - Solve the equation:
     \[
     \frac{5n^2 - 3n + 2}{2} = 2002
     \]
   - Multiply both sides by 2:
     \[
     5n^2 - 3n + 2 = 4004
     \]
   - Rearrange to form a quadratic equation:
     \[
     5n^2 - 3n - 4002 = 0
     \]

3. **Solve the quadratic equation:**
   - Use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 5 \cdot (-4002)}}{2 \cdot 5} = \frac{3 \pm \sqrt{9 + 80040}}{10} = \frac{3 \pm \sqrt{80049}}{10}
     \]
   - Calculate the discriminant:
     \[
     \sqrt{80049} \approx 282.85
     \]
   - Thus:
     \[
     n = \frac{3 \pm 282.85}{10}
     \]
   - The two solutions are:
     \[
     n = \frac{285.85}{10} \approx 28.585 \quad \text{and} \quad n = \frac{-279.85}{10} \approx -27.985
     \]

Since \( n \) must be a natural number, there is no integer solution for \( n \) such that \( P(n) = 2002 \).

The final answer is \( \boxed{ P(3) = 13 } \) and there is no \( n \) such that \( P(n) = 2002 \).