2.4.22. Prove that for any positive sequence $\left\{a_{n}\right\}$, $$ \begin{array}{l} \varliminf_{n \rightarrow \infty} \frac{1}{a_{n}}=\frac{1}{\varlimsup_{n \rightarrow \infty} a_{n}} \\ \varlimsup_{n \rightarrow \infty} \frac{1}{a_{n}}=\frac{1}{\underline{\lim }_{n \rightarrow \infty} a_{n}} \end{array} $$ (Here $\left.\frac{1}{+\infty}=0, \frac{1}{0^{+}}=+\infty\right)$.
What does "$\frac{1}{0^{+}}=+\infty$" in the brackets mean in this context? Why isn't it undefined? If I am proving the case that $\varlimsup_{n \rightarrow \infty} a_{n}$ = $0$, can I say that it is equal to $0^+$ so that I can make the factor to become $+\infty$ ?
In the brackets you see the definitions. That is why it is not undefined. These definitions are standard for the positive extended real number line. (https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations)
It is when $\varlimsup_{n \rightarrow \infty} a_{n}$ = $0$ that the in-context meaning of $\frac{1}{0^{+}}=+\infty$ becomes clear, recalling that your sequence is positive. Your first task is then to show that $$\varliminf_{n \rightarrow \infty} \frac{1}{a_{n}}=+\infty.$$
When $\varlimsup_{n \rightarrow \infty} a_{n}$ = $0$ the notation $0^+$ is appropriate, and makes sense, again keeping in mind that your sequence is positive.
Briefly, that answer to your last question is "yes".