For an integer $n\ge 2$ find all $a_1,a_2,\cdots ,a_n, b_1,b_2,\cdots , b_n$ so that 
(a) $0\le a_1\le a_2\le \cdots \le a_n\le 1\le b_1\le b_2\le \cdots \le b_n;$
(b) $\sum_{k=1}^n (a_k+b_k)=2n;$
(c) $\sum_{k=1}^n (a_k^2+b_k^2)=n^2+3n.$
1. **Initial Setup and Constraints:**
   We are given the following constraints:
   \[
   0 \le a_1 \le a_2 \le \cdots \le a_n \le 1 \quad \text{and} \quad 1 \le b_1 \le b_2 \le \cdots \le b_n
   \]
   \[
   \sum_{k=1}^n (a_k + b_k) = 2n
   \]
   \[
   \sum_{k=1}^n (a_k^2 + b_k^2) = n^2 + 3n
   \]

2. **Analyzing the Sum of \(b_k\):**
   Since \(b_k \ge 1\) for all \(k\), we have:
   \[
   \sum_{k=1}^n b_k \ge n
   \]
   Given \(\sum_{k=1}^n (a_k + b_k) = 2n\), it follows that:
   \[
   \sum_{k=1}^n a_k + \sum_{k=1}^n b_k = 2n
   \]
   Since \(\sum_{k=1}^n b_k \ge n\), we must have:
   \[
   \sum_{k=1}^n a_k \le n
   \]

3. **Analyzing the Sum of Squares:**
   We are given:
   \[
   \sum_{k=1}^n (a_k^2 + b_k^2) = n^2 + 3n
   \]
   Since \(0 \le a_k \le 1\), we have \(a_k^2 \le a_k\). Therefore:
   \[
   \sum_{k=1}^n a_k^2 \le \sum_{k=1}^n a_k
   \]
   Let \(S = \sum_{k=1}^n a_k\). Then:
   \[
   \sum_{k=1}^n a_k^2 \le S
   \]
   Substituting into the sum of squares equation:
   \[
   S + \sum_{k=1}^n b_k^2 = n^2 + 3n
   \]

4. **Bounding \(b_k\):**
   Since \(b_k \ge 1\), we can write:
   \[
   b_k = 1 + c_k \quad \text{where} \quad c_k \ge 0
   \]
   Then:
   \[
   b_k^2 = (1 + c_k)^2 = 1 + 2c_k + c_k^2
   \]
   Summing over all \(k\):
   \[
   \sum_{k=1}^n b_k^2 = \sum_{k=1}^n (1 + 2c_k + c_k^2) = n + 2\sum_{k=1}^n c_k + \sum_{k=1}^n c_k^2
   \]
   Substituting into the sum of squares equation:
   \[
   S + n + 2\sum_{k=1}^n c_k + \sum_{k=1}^n c_k^2 = n^2 + 3n
   \]
   Simplifying:
   \[
   S + 2\sum_{k=1}^n c_k + \sum_{k=1}^n c_k^2 = n^2 + 2n
   \]

5. **Solving for \(a_k\) and \(b_k\):**
   To satisfy all constraints, consider the simplest case where \(a_k = 0\) for all \(k\). Then:
   \[
   S = 0 \quad \text{and} \quad \sum_{k=1}^n a_k^2 = 0
   \]
   The sum of \(b_k\) becomes:
   \[
   \sum_{k=1}^n b_k = 2n
   \]
   The sum of squares of \(b_k\) becomes:
   \[
   \sum_{k=1}^n b_k^2 = n^2 + 3n
   \]
   To satisfy these, let \(b_1 = b_2 = \cdots = b_{n-1} = 1\) and \(b_n = n + 1\). Then:
   \[
   \sum_{k=1}^{n-1} 1 + (n + 1) = n - 1 + (n + 1) = 2n
   \]
   \[
   \sum_{k=1}^{n-1} 1^2 + (n + 1)^2 = n - 1 + (n + 1)^2 = n - 1 + n^2 + 2n + 1 = n^2 + 3n
   \]

Thus, the solution is:
\[
(a_1, a_2, \dots, a_n) = (0, 0, \dots, 0), \quad (b_1, b_2, \dots, b_n) = (1, 1, \dots, 1, n + 1)
\]

The final answer is \( \boxed{ (a_1, a_2, \dots, a_n) = (0, 0, \dots, 0), \quad (b_1, b_2, \dots, b_n) = (1, 1, \dots, 1, n + 1) } \)