If $S$ is a sequence of positive integers let $p(S)$ be the product of the members of $S$. Let $m(S)$ be the arithmetic mean of $p(T)$ for all non-empty subsets $T$ of $S$. Suppose that $S'$ is formed from $S$ by appending an additional positive integer. If $m(S) = 13$ and $m(S') = 49$, find $S'$.
1. Let \( S = \{ a_1, a_2, \ldots, a_n \} \). The arithmetic mean \( m(S) \) of the product of all non-empty subsets of \( S \) is given by:
   \[
   m(S) = \frac{\prod_{i=1}^n (1 + a_i) - 1}{2^n - 1}
   \]
   Given \( m(S) = 13 \), we have:
   \[
   13 = \frac{\prod_{i=1}^n (1 + a_i) - 1}{2^n - 1}
   \]
   This implies:
   \[
   13 \cdot (2^n - 1) = \prod_{i=1}^n (1 + a_i) - 1
   \]
   Therefore:
   \[
   \prod_{i=1}^n (1 + a_i) = 13 \cdot (2^n - 1) + 1
   \]

2. Now, suppose \( S' \) is formed by appending an additional positive integer \( a_{n+1} \) to \( S \). Then:
   \[
   S' = \{ a_1, a_2, \ldots, a_n, a_{n+1} \}
   \]
   Given \( m(S') = 49 \), we have:
   \[
   49 = \frac{\prod_{i=1}^{n+1} (1 + a_i) - 1}{2^{n+1} - 1}
   \]
   This implies:
   \[
   49 \cdot (2^{n+1} - 1) = \prod_{i=1}^{n+1} (1 + a_i) - 1
   \]
   Therefore:
   \[
   \prod_{i=1}^{n+1} (1 + a_i) = 49 \cdot (2^{n+1} - 1) + 1
   \]

3. We know:
   \[
   \prod_{i=1}^{n+1} (1 + a_i) = \left( \prod_{i=1}^n (1 + a_i) \right) (1 + a_{n+1})
   \]
   Substituting the expressions from steps 1 and 2, we get:
   \[
   (13 \cdot (2^n - 1) + 1)(1 + a_{n+1}) = 49 \cdot (2^{n+1} - 1) + 1
   \]

4. Expanding and simplifying:
   \[
   13 \cdot (2^n - 1) \cdot (1 + a_{n+1}) + 1 + a_{n+1} = 49 \cdot (2^{n+1} - 1) + 1
   \]
   \[
   13 \cdot (2^n - 1) + 13 \cdot (2^n - 1) \cdot a_{n+1} + 1 + a_{n+1} = 49 \cdot 2^{n+1} - 49 + 1
   \]
   \[
   13 \cdot 2^n - 13 + 13 \cdot 2^n \cdot a_{n+1} - 13 \cdot a_{n+1} + 1 + a_{n+1} = 98 \cdot 2^n - 49
   \]
   \[
   13 \cdot 2^n + 13 \cdot 2^n \cdot a_{n+1} - 12 \cdot a_{n+1} - 12 = 98 \cdot 2^n - 49
   \]
   \[
   85 \cdot 2^n - 36 = 13 \cdot 2^n \cdot a_{n+1} - 12 \cdot a_{n+1}
   \]

5. Solving for \( a_{n+1} \) and \( n \):
   \[
   85 \cdot 2^n - 36 = 13 \cdot 2^n \cdot a_{n+1} - 12 \cdot a_{n+1}
   \]
   By trying various values of \( a_{n+1} \) and solving for \( n \), we find that \( n = 3 \) and \( a_{n+1} = 7 \) is a solution.

6. Thus, we need to find \( a_1, a_2, a_3 \in \mathbb{Z} \) such that:
   \[
   92 = (1 + a_1)(1 + a_2)(1 + a_3)
   \]
   One possible solution is \( a_1 = 1 \), \( a_2 = 1 \), and \( a_3 = 22 \).

7. Verifying:
   \[
   m(\{1, 1, 22\}) = 13
   \]
   \[
   m(\{1, 1, 22, 7\}) = 49
   \]

Thus, a possible solution for \( S' \) is:
\[
S' = \boxed{\{1, 1, 7, 22\}}
\]