1. **State the problem:** We have two sequences \(\{a_n\}\) and \(\{c_n\}\). We need to find the sum of the first 4 terms of \(\{a_n\}\), denoted \(S_4\), and then find the product of \(S_4\) with the sum of the infinite series of \(\{c_n\}\).\n\n2. **Analyze \(\{a_n\}\):** The sequence is \(-6, 12, -24, 48, \ldots\). This is a geometric sequence where each term is multiplied by \(-2\) to get the next term.\n\n3. **Formula for sum of first \(n\) terms of geometric sequence:**\n$$S_n = a_1 \frac{1-r^n}{1-r}$$\nwhere \(a_1\) is the first term and \(r\) is the common ratio.\n\n4. **Calculate \(S_4\) for \(\{a_n\}\):**\nHere, \(a_1 = -6\), \(r = -2\), and \(n=4\).\n$$S_4 = -6 \times \frac{1 - (-2)^4}{1 - (-2)} = -6 \times \frac{1 - 16}{1 + 2} = -6 \times \frac{-15}{3} = -6 \times (-5) = 30$$\n\n5. **Analyze \(\{c_n\}\):** The sequence is \(-8, 1, -\frac{1}{8}, \frac{1}{64}, \ldots\). This is also geometric with first term \(a_1 = -8\) and common ratio \(r = -\frac{1}{8}\).\n\n6. **Formula for sum of infinite geometric series:**\n$$S_\infty = \frac{a_1}{1-r}$$\nvalid if \(|r| < 1\). Here, \(|-\frac{1}{8}| = \frac{1}{8} < 1\), so the formula applies.\n\n7. **Calculate sum of infinite series for \(\{c_n\}\):**\n$$S_\infty = \frac{-8}{1 - (-\frac{1}{8})} = \frac{-8}{1 + \frac{1}{8}} = \frac{-8}{\frac{9}{8}} = -8 \times \frac{8}{9} = -\frac{64}{9}$$\n\n8. **Find the product:**\n$$S_4 \times S_\infty = 30 \times \left(-\frac{64}{9}\right) = -\frac{1920}{9} = -\frac{640}{3}$$\n\n**Final answer:**\n\[ S_4 = 30 \]\n\[ \text{Product} = -\frac{640}{3} \]