Determine all integers $n$ such that both of the numbers:
$$|n^3 - 4n^2 + 3n - 35| \text{ and }  |n^2 + 4n + 8|$$
are both prime numbers.
To determine all integers \( n \) such that both \( |n^3 - 4n^2 + 3n - 35| \) and \( |n^2 + 4n + 8| \) are prime numbers, we will analyze the given expressions and check for integer solutions.

1. **Analyze the expression \( |n^2 + 4n + 8| \):**
   - We need \( n^2 + 4n + 8 \) to be a prime number.
   - Let's consider the possible values of \( n \) that make \( n^2 + 4n + 8 \) a prime number.

2. **Check for divisibility by 5:**
   - Note that for any integer \( n \), one of the two numbers \( |n^3 - 4n^2 + 3n - 35| \) or \( |n^2 + 4n + 8| \) must be divisible by 5.
   - This is because the polynomial \( n^3 - 4n^2 + 3n - 35 \) modulo 5 will cycle through all residues modulo 5 as \( n \) varies, and similarly for \( n^2 + 4n + 8 \).

3. **Solve \( n^3 - 4n^2 + 3n - 35 = \pm 5 \):**
   - First, solve \( n^3 - 4n^2 + 3n - 35 = -5 \):
     \[
     n^3 - 4n^2 + 3n - 30 = 0
     \]
     - Check for integer solutions:
       - \( n = 1 \):
         \[
         1^3 - 4 \cdot 1^2 + 3 \cdot 1 - 30 = 1 - 4 + 3 - 30 = -30 \neq 0
         \]
       - \( n = 2 \):
         \[
         2^3 - 4 \cdot 2^2 + 3 \cdot 2 - 30 = 8 - 16 + 6 - 30 = -32 \neq 0
         \]
       - Continue checking other integers, but no solutions are found.

   - Next, solve \( n^3 - 4n^2 + 3n - 35 = 5 \):
     \[
     n^3 - 4n^2 + 3n - 40 = 0
     \]
     - Check for integer solutions:
       - \( n = 5 \):
         \[
         5^3 - 4 \cdot 5^2 + 3 \cdot 5 - 40 = 125 - 100 + 15 - 40 = 0
         \]
         - This is a solution. Check if \( n^2 + 4n + 8 \) is prime:
           \[
           5^2 + 4 \cdot 5 + 8 = 25 + 20 + 8 = 53
           \]
           - 53 is a prime number.

4. **Solve \( n^2 + 4n + 8 = \pm 5 \):**
   - First, solve \( n^2 + 4n + 8 = -5 \):
     \[
     n^2 + 4n + 13 = 0
     \]
     - The discriminant is:
       \[
       \Delta = 4^2 - 4 \cdot 1 \cdot 13 = 16 - 52 = -36
       \]
       - No real solutions.

   - Next, solve \( n^2 + 4n + 8 = 5 \):
     \[
     n^2 + 4n + 3 = 0
     \]
     - The discriminant is:
       \[
       \Delta = 4^2 - 4 \cdot 1 \cdot 3 = 16 - 12 = 4
       \]
       - The solutions are:
         \[
         n = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2} = -1 \text{ or } -3
         \]
       - Check if \( n = -1 \) makes \( |n^3 - 4n^2 + 3n - 35| \) prime:
         \[
         |-1^3 - 4 \cdot (-1)^2 + 3 \cdot (-1) - 35| = |-1 - 4 - 3 - 35| = |-43| = 43
         \]
         - 43 is a prime number.
       - Check if \( n = -3 \) makes \( |n^3 - 4n^2 + 3n - 35| \) prime:
         \[
         |-3^3 - 4 \cdot (-3)^2 + 3 \cdot (-3) - 35| = |-27 - 36 - 9 - 35| = |-107| = 107
         \]
         - 107 is a prime number.

Hence, the integers \( n \) that satisfy the conditions are \( n = -3, -1, 5 \).

The final answer is \(\boxed{n \in \{-3, -1, 5\}}\).