This is easiest to think about if you have a 'simple' null hypothesis and a
simple alternative. Suppose your data is summarized as a sample mean $\bar X$
of $n = 16$ independent observations from $\mathsf{Norm}(\mu, \sigma=4)$
and you are testing $H_0: \mu = 20$ vs. $H_1: \mu = 30.$
Then the test statistic is
$$Z = \frac{\bar X - 20}{\sigma/\sqrt{n}} = \bar X - 20.$$
Then, under $H_0,$ we have $Z \sim \mathsf{Norm}(0,1).$ Testing at significance
level $\alpha = 0.05 = 5\%,$ you would reject $H_0$ if $Z \ge 1.645.$
That is to say $P(\text{Reject} H_0\,|\,H_0 \text{True}) = P(\text{Type I Error}) = \alpha = 0.05.$ So if 100 students are performing this test and if $\mu = 20,$ then you would expect
five of them to Reject even though $H_0$ is true. This is the answer to part (b).
In my contrived simple example, there is only one way for $H_0$ to be false and that is to have $\mu = 30.$ Then it is easy to find
$$P(\text{Reject } H_0\,|\, H_0 \text{ False}) = P(\text{Reject } H_0\,|\, \mu = 30) = P(Z = \bar X - 20 \ge 1.645\,|\,\mu = 30).$$ That is the "power of the test" sometimes denoted $\gamma = 1 - P(\text{Tyee II Error}) = 1 - \beta.$ You could use $\beta$ to answer part (a).
In practical applications, the alternative hypothesis usually has many
values: for example, $H_1: \mu > 20.$ Then the power and the probability of
Type II Error are functions of the possible alternative values of $\mu,$
sometimes written as $\gamma(\mu_1)$ and $\beta(\mu_1),$ where $\mu_1 \in H_1.$
In such a situation, the power function and the Type II error function have to be evaluated one point at a time.
Accordingly, there would be multiple answers to part (a).
