Significant figure rules In a simple physics experiment, we take the average a few readings to reduce the random errors. I apply significant figure rules to these.
Say we round off at each step:
(8.0+9.0+10.0)/3 = 27.0/3 (addition keeps dp) = 9.00 (division keeps sf)
Say we choose to only round off at the end by taking the least precise of given readings:
(9.0+10.0+11.0)/3 = 30.0/3=10 (2sf because 9.0 is 2 sf
'common sense' tells us that in the first example, the decimal place should be kept at 9.0, and in the second, it should be 10.0, not 10. 
Another example: Given the external (19.2) and internal (18.3) diameter of a pipe, calculate its thickness: (19.2-18.3)/2 =0.45 or 0.5 or 0.450?
All these significant figure rules seem to only work out in theory to me. Can someone explain?
 A: I sympathise. I don't like s.f. as they really are just a very rough approximation of errors. If you have worry about a rounding method, you can always work out the real error to find what the real rounding should be. 
Just to repeat your rule, addition/subtraction, work with decimal places, multiplication use sig figs. This translates to addition/subtraction add absolute error, multiplication/division add percentage error.  
example 1
Use standard error handling:
(9.0+10.0+11.0)/3 = (30.0 +/- 0.15)/3 = 10.0 +/- 0.05
A problem here is that using plain d.p. or s.f. doesn't add errors properly, so 30.0 looks more accurate than it really is. It is really +/- 0.15 not +/- 0.05.
The main problem with only using 2 sig figs for 9.0 to 10 is that 9.0 +/- 0.05 has under 0.6% error, while 10 +/- 0.5 is 5% error. It makes more sense to use 10.0 which has a similar 0.5% error. We have a seeming paradox where 9.0 is 2 sig figs, has almost the same accuracy as 10.0 which looks like 3 sig figs. Leading to:
my extra rule 
When the first number of a value is a one, I count the significant figures that come after the '1'. So all of the original numbers in your example are really just 2 s.f.  This means 1.1 is really just 1 significant figure, the same as 9.
Standard deviation error handling It is possible to use standard deviation error handling rather than extreme error regime (add squares of errors, then take sqrt). The above would be:
(9.0+10.0+11.0)/3 = (30.0 +/- 0.09)/3 = 10.0 +/- 0.03
Due to laws of statistics, the more items you average the more accurate the mean is, even if the individual measurements are not very accurate. But in this case it gives basically the same result as normal error handling. 
Another example: Given the external (19.2) and internal (18.3) diameter of a pipe, calculate its thickness: 
(19.2-18.3)/2 = (0.9 +/- 0.1)/2 = 0.45 +/- 0.05 = 0.5 (1s.f.!)
As you have discovered, finding differences is a great way of destroying any s.f. you might have started out with!
Using standard deviation error handling, the final answer is 0.45 +/- 0.04, and there is not really a good way of expressing this using plain rounding, as neither 0.5 or 0.4 express the range of possible value very well, while 0.45 looks like it is +/- 0.005, which is nearly 10 times more accurate than it really is. 
I would just stick with 0.5.
[edit] I would only use the simplified 0.5 for mental approximation, communicating to a non-physicist or under duress, and not in real physics work. If I was at all able to express it in proper error language I would do that: 0.45 +/- 0.04. There will always be problems and trade offs when you are attempting to express 2 pieces of information (measurement and error) using a single number. 
A: I think it is difficult to make general rules, but you should learn to recognise what is right and wrong.
When you are averaging lots of numbers with the same precision, it is perfectly possible to arrive at and quote a mean that is more precise than any of the original readings. But the precision only increases as $\sqrt{n}$ (actually $\sqrt{n-1}$ if using the standard deviation of the population to estimate the error in the mean).
In your first examples I take it that the precision of each reading is $\pm 0.05$ (actually that is bigger than the standard deviation, because presumably it is more like a 95% confidence interval, but let us proceed). The standard error combination formulae tells us that the uncertainty in the mean is $\sqrt{3 \times 0.05^2}/3 = 0.029$. In this case I would quote the result as $10.00 \pm 0.03$. Note that when you quote a number like this there is actually no rule that says the central value has to be the most likely one; just that there is a defined confidence that it lies between the upper and lower error bar.
I don't like to see any mismatch in the decimal places in floating point numbers and their errors. $10.0 \pm 0.03$ looks bizarre to me and in the examples you have quoted it is the decimal places that matter, not the significant figures.
Second example - the diameter of the pipe is (assuming that the diameters are what is actually measured) $(19.2 \pm 0.05 -  18.3 \pm 0.05)/2 = 0.45 \pm 0.04$ (and you would quote this, with its uncertainty). The uncertainty here comes by adding the individual errors in quadrature and dividing by 2 (it's actually 0.0353, but I would always err on the side of rounding up an error like this, even if it had been 0.034). I would not use 0.5 as the answer simply because, whilst a plausible value, it is (just) outside the confidence interval of your measurements.
Much of this is verging on philosophy unless you specify (or know) exactly what the probability distribution of your uncertainties is. And never just quote a number as a result without attaching an uncertainty - as you have done in your question.
