This seems to be something like a self-study exercise so I won't presently give a full solution to every part but I will get you a fair way along.
Basic results on expectations and variances:
$\qquad$1. $ E(\sum_i s_i) = \sum_i E(s_i)$
$\qquad$2. $ \text{Var}(\sum_i s_i) = \sum_i \text{Var}(s_i), \:$ for independent variates
For iid exponentials with mean $\mu$, for given $C_j$, $S_j$ has population mean $C_j\, \mu$ and $S_j/C_j = \overline{s}$ has population mean $\mu$. You could estimate $\mu$ from a single $(S,C)$ pair. If you have more than one of them, you will want to weight them appropriately in a weighted average.
For iid normals with mean $\mu$ and standard deviation $\sigma$, for given $C_j$, $S_j$ has population mean $C_j\, \mu$ and $S_j/C_j = \overline{s}$ has population mean $\mu$. The variance of $S_j$ is $C_j\, \sigma^2$, and the variance of $\bar{s}$ is $\sigma^2/C_j$. If you only have $S$ and $C$ values available to you, you'll require more than one of them to estimate $\sigma$.
Usually, you'd solve problems like this via the method of maximum likelihood, rather than just by equating moments, but in these two cases, that turns out to be essentially the same thing.