Multiplying means and calculating variance 
*

*I have multiplied together two means and now want to calculate the overall standard deviation. The two means and standard deviation are here: 13.7 +/- 12.7 (1SD) and 4.0 +/- 2.6 (1SD). So the answer would be 55.11 with what overall standard deviation?

*I then want to multiply this answer of 55.1 by 92. There is no standard deviation attached to this because it is 92 days. Does the standard deviation stay the same as the answer for question 1?   
Edit: My overall goal is looking at seed dispersal in the dry season. There are around 13 seeds in a poo and they poo around 4 times a day. That tells me the number of seeds dispersed on average per day. By multiplying that by 92 it gives a rough indication of the number of seeds dispersed across the dry season. This hasn't previously been studied before, it's not designed to be overly accurate but demonstrate the importance of mutualism in seed dispersal.
 A: I found some old notes on this which I think come from the page I linked to in comments above http://www.fas.harvard.edu/~scphys/nsta/error_propagation.pdf so any intellectual credit goes to the authors of that set of notes. The first question of the OP is answered I believe in the section Multiplication (or division) and the second in the section Multiplication by a constant.
If you measure some quantities $a, b, c, \dots$ with uncertainties
$\delta{a}, \delta{b}, \delta{c}, \dots$
what happens when you combine them?
Addition (or subtraction)
If
\begin{equation}
Q = a + b + \dots + c - (x + y + \dots + z)
\end{equation}
then
\begin{equation}
\delta{Q} = \sqrt{(\delta{a})^2 + (\delta{b})^2 + \dots + (\delta{c})^2 + (\delta{x})^2 + (\delta{y})^2 + \dots + (\delta{z})^2}
\end{equation}
Multiplication (or division)
If
\begin{equation}
Q = \frac{ab\dots{}c}{xy\dots{}z}
\end{equation}
then
\begin{equation}
\frac{\delta{Q}}{\vert{}Q\vert} = \sqrt{\left(\frac{\delta{a}}{a}\right)^2 +
\left(\frac{\delta{b}}{b}\right)^2 +
\dots +
\left(\frac{\delta{c}}{c}\right)^2 +
\left(\frac{\delta{x}}{x}\right)^2 +
\left(\frac{\delta{y}}{y}\right)^2 +
\dots +
\left(\frac{\delta{z}}{z}\right)^2}
\end{equation}
Multiplication by a constant
If $A$ is known exactly and
\begin{equation}
Q = Ax
\end{equation}
then
\begin{equation}
\delta{Q} = \vert{}A\vert\delta{x}
\end{equation}
A: Variance of a product of two random variables is discussed in the post Variance of product of dependent variables. This is for the general case of dependent scenario and the independent case can be obtained by setting the covariance zero. 
Specifically using the result $${\rm Var}(XY)= {\rm cov}(X^{2}, Y^{2}) + [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(Y) + E(Y)^2] - [ {\rm cov}(X,Y) + E(X)E(Y) ]^{2}$$ from the post, you can replace $X$ and $Y$ respectively with $\bar X$ and $\bar Y$. Then taking the square root will get the standard deviation of $\bar X \times \bar Y$. 
Then, my suggestion would be to use the expression and consider a "plug-in" estimator. Note that this cannot be computed knowing only the standard deviations, but the computation (with some data manipulation) is possible if the raw data is available. 
