Determine all functions $f : \mathbb N_0 \rightarrow \mathbb N_0 - \{1\}$ such that
\[f(n + 1) + f(n + 3) = f(n + 5)f(n + 7) - 1375, \qquad \forall n \in \mathbb N.\]
1. Given the functional equation:
   \[
   f(n + 1) + f(n + 3) = f(n + 5)f(n + 7) - 1375, \qquad \forall n \in \mathbb{N}.
   \]
   We need to determine all functions \( f : \mathbb{N}_0 \rightarrow \mathbb{N}_0 - \{1\} \) that satisfy this equation.

2. Consider the equation for \( n \) and \( n+2 \):
   \[
   f(n + 1) + f(n + 3) = f(n + 5)f(n + 7) - 1375,
   \]
   \[
   f(n + 3) + f(n + 5) = f(n + 7)f(n + 9) - 1375.
   \]

3. Subtract the first equation from the second:
   \[
   (f(n + 3) + f(n + 5)) - (f(n + 1) + f(n + 3)) = (f(n + 7)f(n + 9) - 1375) - (f(n + 5)f(n + 7) - 1375).
   \]
   Simplifying, we get:
   \[
   f(n + 5) - f(n + 1) = f(n + 7)(f(n + 9) - f(n + 5)).
   \]

4. Similarly, consider the equation for \( n+2 \) and \( n+4 \):
   \[
   f(n + 5) + f(n + 7) = f(n + 9)f(n + 11) - 1375,
   \]
   \[
   f(n + 7) + f(n + 9) = f(n + 11)f(n + 13) - 1375.
   \]

5. Subtract the first equation from the second:
   \[
   (f(n + 7) + f(n + 9)) - (f(n + 5) + f(n + 7)) = (f(n + 11)f(n + 13) - 1375) - (f(n + 9)f(n + 11) - 1375).
   \]
   Simplifying, we get:
   \[
   f(n + 9) - f(n + 5) = f(n + 11)(f(n + 13) - f(n + 9)).
   \]

6. By induction, we can generalize:
   \[
   f(n + 5) - f(n + 1) = f(n + 7)f(n + 11) \cdots f(n + 4k - 1)(f(n + 4k + 1) - f(n + 4k - 3)),
   \]
   for any positive integer \( k \).

7. If \( f(n + 5) - f(n + 1) \neq 0 \), then none of \( f(n + 7), f(n + 11), \ldots, f(n + 4k - 1), (f(n + 4k + 1) - f(n + 4k - 3)) \) may equal 0. Since \( f(n) \neq 1 \), we have \( f(n + 7), f(n + 11), \ldots, f(n + 4k - 1) \geq 2 \) and \( f(n + 4k + 1) - f(n + 4k - 3) \geq 1 \).

8. Therefore, \( f(n + 5) - f(n + 1) \geq 2^{k - 1} \) for any positive integer \( k \), which is impossible. Hence, \( f(n + 5) - f(n + 1) = 0 \), i.e., \( f(n + 5) = f(n + 1) \).

9. This implies that \( f \) is periodic with period 4. Let \( f(4) = a, f(5) = b, f(2) = c, f(3) = d \). We have:
   \[
   a + c = ac - 1375,
   \]
   \[
   b + d = bd - 1375.
   \]

10. Solving these equations, we get:
    \[
    ac - a - c = 1375,
    \]
    \[
    bd - b - d = 1375.
    \]

11. Rearranging, we get:
    \[
    (a - 1)(c - 1) = 1376,
    \]
    \[
    (b - 1)(d - 1) = 1376.
    \]

12. The factors of 1376 are \( 1, 2, 4, 8, 16, 32, 43, 86, 172, 344, 688, 1376 \). Thus, \( a, c, b, d \) can be any pairs such that \( (a - 1)(c - 1) = 1376 \) and \( (b - 1)(d - 1) = 1376 \).

13. Therefore, the function \( f \) can be any function such that \( f(n) \) is periodic with period 4 and satisfies the above conditions.

The final answer is \( \boxed{ f(n) } \) is periodic with period 4 and satisfies \( (f(n+4k+1) - 1)(f(n+4k+3) - 1) = 1376 \) for all \( k \in \mathbb{N}_0 \).