a) Find all integer pairs $(x,y)$ satisfying $x^4+(y+2)^3=(x+2)^4.$

b) Prove that: if $p$ is a prime of the form $p=4k+3$ $(k$ is a non-negative number$),$ then there doesn's exist $p-1$ consecutive non-negative integers such that we can divide the set of these numbers into $2$ distinct subsets so that the product of all the numbers in one subset is equal to that in the remained subset.
### Part (a)
We need to find all integer pairs \((x, y)\) satisfying the equation:
\[ x^4 + (y+2)^3 = (x+2)^4. \]

1. **Expand the right-hand side:**
   \[ (x+2)^4 = x^4 + 4x^3 \cdot 2 + 6x^2 \cdot 4 + 4x \cdot 8 + 16 = x^4 + 8x^3 + 24x^2 + 32x + 16. \]

2. **Rewrite the equation:**
   \[ x^4 + (y+2)^3 = x^4 + 8x^3 + 24x^2 + 32x + 16. \]

3. **Subtract \(x^4\) from both sides:**
   \[ (y+2)^3 = 8x^3 + 24x^2 + 32x + 16. \]

4. **Consider the polynomial on the right-hand side:**
   \[ (y+2)^3 = 8(x^3 + 3x^2 + 4x + 2). \]

5. **Since \((y+2)^3\) must be a perfect cube, we need to check if \(8(x^3 + 3x^2 + 4x + 2)\) can be a perfect cube.**

6. **Check for small integer values of \(x\):**
   - For \(x = 0\):
     \[ (y+2)^3 = 16 \implies y+2 = 2 \implies y = 0. \]
     Thus, \((x, y) = (0, 0)\) is a solution.

   - For \(x = 1\):
     \[ (y+2)^3 = 8(1 + 3 + 4 + 2) = 80, \]
     which is not a perfect cube.

   - For \(x = -1\):
     \[ (y+2)^3 = 8(-1 + 3 - 4 + 2) = 0 \implies y+2 = 0 \implies y = -2. \]
     Thus, \((x, y) = (-1, -2)\) is a solution.

7. **Verify if there are any other solutions:**
   - For \(x = 2\):
     \[ (y+2)^3 = 8(8 + 12 + 8 + 2) = 240, \]
     which is not a perfect cube.

   - For \(x = -2\):
     \[ (y+2)^3 = 8(-8 + 12 - 8 + 2) = -16, \]
     which is not a perfect cube.

Thus, the only integer solutions are \((x, y) = (0, 0)\) and \((x, y) = (-1, -2)\).

### Part (b)
We need to prove that if \(p\) is a prime of the form \(p = 4k + 3\), then there do not exist \(p-1\) consecutive non-negative integers such that we can divide the set of these numbers into 2 distinct subsets so that the product of all the numbers in one subset is equal to that in the remaining subset.

1. **Assume there exist \(p-1\) consecutive non-negative integers \(a_1, a_2, \ldots, a_{p-1}\) such that we can divide the set into two subsets with equal product.**

2. **Case 1: There exists \(a_i \equiv 0 \pmod{p}\):**
   - This would imply that one subset contains \(0\), making the product of that subset \(0\), which cannot equal the product of the other subset unless it also contains \(0\), leading to a contradiction.

3. **Case 2: All \(a_i \not\equiv 0 \pmod{p}\):**
   - By Wilson's Theorem, \((p-1)! \equiv -1 \pmod{p}\).
   - The product of all \(a_i\) modulo \(p\) is \((p-1)! \equiv -1 \pmod{p}\).

4. **Since we can divide the set into two subsets with equal product, let the products of the two subsets be \(A\) and \(B\) such that \(A \cdot B \equiv -1 \pmod{p}\).**

5. **This implies \(A^2 \equiv -1 \pmod{p}\):**
   - Therefore, \(A^4 \equiv 1 \pmod{p}\).
   - The order of \(A\) modulo \(p\) must be \(4\), which divides \(p-1 = 4k + 2\), leading to a contradiction since \(4\) does not divide \(4k + 2\).

Thus, no such division is possible, and the statement is proven.

\(\blacksquare\)

The final answer is \( \boxed{ (x, y) = (0, 0) } \) and \((x, y) = (-1, -2)\).