I'm very scared that my calculations I did were wrong. Here is why: I assumed true standard deviation meant population S.D. However the question says the standard deviation is from a sample. So what exactly is true standard deviation?
I did a quick google search which gave me standard error. Isn't standard error the standard deviation of the sampling distribution? We are only given a single sample.
The second text highlighted is simply to verify if anyone else got the answer 4.52-5.18.
In my experience the expression 'true standard deviation' is often used to mean 'population standard deviation', as distinguished from 'estimated standard deviation' or 'sample standard deviation'.
Then, in the formula $\bar X \pm 1.96\sigma/\sqrt{n},$ you have $\bar X = 4.85,\; \sigma = 0.75$ and $n = 20.$ This gives the CI $(4.521, 5.179)$ in agreement with your result. (Because $\bar X$ and $\sigma$ are given to two places, some authors would suggest retaining three places in the endpoints of the CI.)
Note: If the given SD were intended to be the sample standard deviation $s = 0.75,$ then you would use 2.093 from a t table with $n-1=19$ degrees of freedom. But in that case the problem should have said 'sample SD'. Also, note that parts (c) and (d) ask you to find $n,$ in which case you would not know what DF to use, and you would need to work at a deeper level than suggested by the elementary drill nature of this problem