Find the triplets of primes $(a,\ b,\ c)$ such that $a-b-8$ and $b-c-8$ are primes.
To find the triplets of primes \((a, b, c)\) such that \(a - b - 8\) and \(b - c - 8\) are primes, we start by analyzing the given conditions.

1. **Initial Conditions and Constraints:**
   - Since \(a - b - 8\) and \(b - c - 8\) are primes, it implies that \(a > b > c \geq 2\).
   - Clearly, \(a\) and \(b\) must be odd primes because if either were even, the difference would not be a prime (except for the prime number 2, but that would not fit the constraints).

2. **Analyzing \(a - b - 8\):**
   - Since \(a\) and \(b\) are odd, \(a - b\) is even.
   - Therefore, \(a - b - 8\) is even. The only even prime is 2.
   - Thus, \(a - b - 8 = 2\) which simplifies to \(a = b + 10\).

3. **Considering \(c = 2\):**
   - If \(c = 2\), then \(b - c - 8 = b - 10\) must be a prime.
   - Therefore, \(b - 10\) must be a prime.
   - Since \(b\) is an odd prime, \(b - 10\) must also be an odd prime.

4. **Modulo 3 Analysis:**
   - Under modulo 3, we have:
     - \(a = b + 10\)
     - \(b\)
     - \(b - 10\)
   - This implies:
     - \(a \equiv b + 1 \pmod{3}\)
     - \(b \equiv b \pmod{3}\)
     - \(b - 10 \equiv b - 1 \pmod{3}\)
   - Therefore, at least one of \(a\), \(b\), or \(b - 10\) must be divisible by 3.

5. **Finding \(b\):**
   - For \(b - 10\) to be a prime and divisible by 3, it must be 3.
   - Thus, \(b - 10 = 3\) which gives \(b = 13\).
   - Therefore, \(a = b + 10 = 23\).

6. **Triplet for \(c = 2\):**
   - The triplet is \((a, b, c) = (23, 13, 2)\).

7. **Considering \(c\) is an odd prime greater than 3:**
   - Since \(b\) and \(c\) are odd, \(b - c - 8 = 2\) which implies \(b = c + 10\).
   - Therefore, \(a = b + 10 = c + 20\).

8. **Modulo 3 Analysis for \(c\):**
   - Under modulo 3, we have:
     - \(a = c + 20\)
     - \(b = c + 10\)
     - \(c\)
   - This implies:
     - \(a \equiv c - 1 \pmod{3}\)
     - \(b \equiv c + 1 \pmod{3}\)
     - \(c \equiv c \pmod{3}\)
   - Therefore, at least one of \(a\), \(b\), or \(c\) must be divisible by 3.

9. **Finding \(c\):**
   - For \(c\) to be a prime and divisible by 3, it must be 3.
   - Thus, \(c = 3\).
   - Therefore, \(b = c + 10 = 13\) and \(a = c + 20 = 23\).

10. **Triplet for \(c = 3\):**
    - The triplet is \((a, b, c) = (23, 13, 3)\).

In all cases, the triplets of primes such that \(a - b - 8\) and \(b - c - 8\) are primes are \((a, b, c) = (23, 13, 2)\) and \((23, 13, 3)\).

The final answer is \(\boxed{(23, 13, 2)}\) and \(\boxed{(23, 13, 3)}\).