I assume the sample is supposed to be independent. I denote the content of the 16 sampled bottles by $X_1,X_2,\ldots,X_{16}$.
Regarding your first question: The variance of the sample mean is
$$\operatorname{Var}(\bar{X})=\operatorname{Var}\left(\frac{1}{16}\sum_{i=1}^{16}{X_i}\right)=\frac{1}{16^2}\operatorname{Var}\left(\sum_{i=1}^{16} {X_i}\right)=\frac{1}{16^2}\sum_{i=1}^{16}\operatorname{Var}(X_i)=\frac{1}{16^2} 16\cdot 0.02^2 = \frac{0.02^2}{16},$$ where we have used independence and the rules for the variance.
The same reasoning applies to (b). By the same steps as above (just replace 16 by $n$) you can show that the variance of the sample mean is equal $\frac{\sigma^2}{n}$ and hence the standard deviation is $\frac{\sigma}{\sqrt{n}}$. Note that the variance of the sample mean does not equal to the variance of the underlying distribution (except if $n=1$ or the variance of the underlying distribution is $0$).
The expected value on the other hand is the same, which the following calculation shows:
$$\mathbb{E}(\bar{X})=\mathbb{E}\left(\frac{1}{n}\sum_{i=1}^{n}{X_i}\right)=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left(X_i\right)=\frac{1}{n}n\cdot0.99=0.99,$$ where we have used the linearity of the expectation.
That's how your prof got the expected value and the standard deviation for the sample average.
To think about this intuitively, imagine we collect a very small sample (so $n$ is small). Each observation carries much weight in the sample average (especially unusually small or large observations). If we draw repeated small samples, our sample average will hence vary a lot. If we have a very large sample ($n$ large), then we gonna observe all kinds of observation (most of them in a range that has much probability mass) and a few extreme observations won't move the sample average by much. If we draw repeated large samples, our sample average will not vary a lot. So, we would intuitively assume that the variance decreases as the sample size increases (indeed $n$ shows up in the denominator).
