Comparison of 2 distributions I have 2 distributions, 1 is, as far as im aware, normally distributed. Distribution 1 is the control group.
1st distribution
Mean = 0.000002757; Median = 0; StDev = 0.00119307; Number of data points = 91601
2nd distribution
Mean = 0.000058; Median = 0.000125; StDev = 0.001646243; Number of data points = 94045
(Question 1) Im assuming i can do a hypothesis test of the difference of the means to determine if the 2nd distribution is significantly different to the first? 
1) H0 = No significant change.

2) Pick 5% significance level.

3) Z-table gives 1.65.

4) Calc Stdev of distribution of the difference of the means = 6.66*10^-6

5) Find threshold value =>(6.6610^-6)*1.65 = 1.098*10^-5

Since the difference of the difference of the means > 1.098*10^-5 i can reject the null hypothesis?
(Question 2) Does it matter that the sample sizes are slightly different, regarding the calc for step 4?
(Question 3) If my 1st distribution was not normal, could i still perform the same calculation to determine if the 2nd distribution was significantly different?
 A: To compare the two distributions, you can do a Kolmogorvo-Smirnov test. The R function ks.test does it : http://stat.ethz.ch/R-manual/R-patched/library/stats/html/ks.test.html 
ks.test(distrib1,distrib2)

The two samples do not have to have the same size.
A: I haven't checked your calculations, but assuming those are right, you should be okay.
You don't need normality - your sample sizes are large enough that you should be able to treat your sample means as normally distributed with the sample variance used as the population variance. (If the populations are strongly non-normal $-$ as they would be if the measurements are necessarily positive, given how much larger than the means the standard deviations are $-$ then power could be an issue, but again the sample sizes are quite large.)
Different sample size doesn't prevent you from doing this.

Say my distributions were gathered from testing steels with different additives. Distribution 2 = steel with additive A and Distribution 3 = steel with additive B. Could i use the difference of their means from distribution 1 as a measure of which is better?

Where distribution 1 is the control? Well, it's possible, but it may not be of any use. You have to take account of the dependence between those two differences - the variance of the difference of differences has to take into account the presence of the first mean. The estimate of the difference of differences itself will just be the difference of the means of the samples from Distributions 2 and 3 - you end up with the same estimate of difference, and (if you do it right) the same variance of the difference. You gain nothing by involving data from distribution 1 except some additional calculation which removes its effect. If you can, it's generally easier to do the comparison you want.
If you have three such distributions, you may want to consider combining them into an analysis of variance type of structure (it's not critical to do so).
A: You could also do a QQPlot to see if the two distribtuions are different from each other.  Here is the R Documentation on how to compare two datasets to see if they are distributed the same. 
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/qqnorm.html
The plot will look as a straight line if the two distributions are the same.  Any deviation will be represented by a deviance from a straight line.
A: If you want a quantitative comparison (a number telling you \emph{how much} are the distributions different) you could also look at the Kullback-Leibler Divergence between $p$ and $q$, i.e. $$D(p||q) = \sum_{x_i} \log \frac{p(x_i)}{q(x_i)} p(x_i)$$
in the discrete case, measured in bits. A similar metric applies to the continuous case (check Wikipedia). Please note that it is not symmetric, i.e. $D(p||q) \neq D(q||p)$.
If the distributions are empirical, i.e. estimated using a histogram, then $D(p||q)$ will have a floor value determined by the estimator.
