Significance levels, mean testing Slabs of toffee are specified by a manufacturer to have a mean weight
of 200 g and the standard deviation is known to be 5g. Weights are
assumed Normally distributed. 

It is found that two samples, each of 100 bars, have mean weights 199.2
g and 199.3 g. Test each mean separately and also their sum to see
whether or not they are significantly less at the 5% level than their
expected values according to the manufacturer’s specification of 200g.
Which of the three results is the most important and why?

I don't seem to understand what they mean by 'test each mean'. What do I have to find?
 A: Imagine  taking a sample of size $100$. The average of their weights is a random variable $\bar{X}$.
Imagine taking another sample of size $100$. The average of their weights is a random variable $\bar{Y}$.
Imagine taking the two samples above, and pooling them to make a sample of $200$. Their mean weight is a random variable $\bar{W}$, which in this case is $\dfrac{\bar{X}+\bar{Y}}{2}$.
First look at $\bar{X}$. This is a random variable. If the true mean of a slab is $200$ with standard deviation $5$, then the random variable $\bar{X}$ has mean $200$ and standard deviation $\dfrac{5}{\sqrt{100}}=0.5$.
Also, if the sample is random, and weights are normally distributed, then $\bar{X}$ has normal distribution.
You want to find the probability that a normal with mean $200$ and standard deviation $0.5$ is less than $199.2$. Note that $199.2$ is $0.8$ grams below the supposed mean of $\bar{X}$. In terms of "standard deviation units," the result is $1.6$ standard deviation units below the mean. 
The probability of this is the probability that a standard normal is less than $-1.6$. You can find the probability of that from a table of the standard normal. If the result is less than $0.05$, then at the $5\%$ significance level we have indication of a problem. The table should give you approximately $0.0548$. So at the $5\%$ significance level we cannot claim that there is a problem: the "low" average might be due to normal fluctuation.
Repeat with the other sample. Again you will not be able to claim, at significance level $5\%$, that there is an issue with the weights. 
For the pooled samples, the experimental sample mean is now $199.25$. And very importantly, the standard deviation of a random variable $\bar{W}$ that comes from a sample of $200$ is $\dfrac{5}{\sqrt{200}}$. Repeat the calculation, and decide whether at the $5\%$ significance level, there is indication of a problem.  You will find that being $0.75$ or more below the mean in a sample of $200$ is highly unlikely, and definitely a big issue at the $5\%$ significance level. 
