The binomial distribution is a distribution commonly used to model repeated experiments with a clear success/fail where each trial is independent of one another and each trial has the same chance of success. For example, when flipping an unfair coin $n$ times and counting the number of heads you see, or selecting balls from a bag with replacement.
Here for our problem, we need to make the assumption that the number of students we are selecting from is a "large" number compared to $300$. In doing so, by modelling the process by a binomial distribution, we can get a good estimates to our probability calculations which are accurate to several decimal points of precision. The idea being, with so many people the difference between whether we were to have selected with replacement or without is so minuscule that it doesn't affect our probability by a significant amount.
Letting $X$ be the random variable which counts the number of successes, the probability of getting exactly $k$ successes out of $n$ independent trials where chance of success is $p$ for each trial is:
$$Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$
By plugging in $p=0.2, n=300$ and $k=50$ to the above, we get an immediate answer for (iii) of
$$Pr(X=50)=\binom{300}{50}\cdot 0.2^{50}\cdot (1-0.2)^{300-50}\approx 0.0207557$$
Now, for the earlier parts, if we have a strong enough calculator, we could just loop through a sum and get an exact answer such as this, but I expect that the intended solution is instead to approach using the normal approximation to the binomial.
It is well known that a binomially distributed random variable has average $np$ and standard deviation $\sqrt{np(1-p)}$, so in our case we have an average of $60$ and a standard deviation of $\sqrt{48}\approx 6.93$.
You may approximate the discrete binomial distribution for given values of $n,p$ by using the continuous normal distribution with mean $np$ and standard deviation $\sqrt{np(1-p)}$, the areas under the graph for each over a region will be very similar so long as $n$ and $np(1-p)$ are large enough. The larger they are, the more accurate the approximation will be.
To continue, to find the approximation for $Pr(a\leq X\leq b)$, we first apply a slight continuity correction by padding the distance away from the region we are interested in for each bound by adding or subtracting $\frac{1}{2}$ so we look instead at $Pr(a-\frac{1}{2}\leq X\leq b+\frac{1}{2})$. This, again, because the binomial distribution was discrete while the normal distribution is continuous. (With large enough standard deviation, this step might even be skipped)
Next, we find "how many standard deviations above the mean $a-\frac{1}{2}$ and $b+\frac{1}{2}$ are respectively (a negative value corresponding to actually being that amount below the mean). Through a bit of algebra, you will find that they are $\dfrac{a-\frac{1}{2}-np}{\sqrt{np(1-p)}}$ and $\dfrac{b+\frac{1}{2}-np}{\sqrt{np(1-p)}}$ respectively. (You may see this instead written as $\frac{a-\mu}{\sigma}$.) You will commonly hear these referred to as "$Z$-Scores" or "Standard Scores".
Now that we have found those values, we can ask the question for a standard normal distribution, asking what the probability is that it lies within that range of standard deviations away from those values. This can be done by looking at a table or using a calculator.
When using a table, remember that $Pr(m\leq Z\leq n) = Pr(Z\leq n) - Pr(Z< m)$
For the example of finding the probability between $20$ and $60$ people are English majors, we go through the effort of calculating the appropriate z-scores, in this case being $\approx\dfrac{19.5-60}{6.93}\approx -5.844$ and $\approx\dfrac{60.5-60}{6.93}\approx 0.072$ respectively.
Finding the probabilities of these, you can look on a table to find $Pr(Z\leq -5.844)\approx 0.0000$ and $Pr(Z\leq 0.072)\approx 0.5287$ respectively, so the probability of being between $20$ and $60$ people is approximately $0.5287$ (compare to the exact answer of $\approx 0.534476$ we found earlier with exact calculation). (compare also to the answer of $0.5$ that we would have gotten had we not used the continuity correction).
The other part can be calculated similarly.
