Interesting: This problem is sort of backwards from the usual ones, where you're given the specificity and sensitivity of the test, and the true prevalence of the disease, and you're expected to obtain the probability that a positive result is a true positive.
Here, instead, you're given the specificity ($95$ percent) and sensitivity ($99$ percent) as usual, but you're given the probability that a positive result is a true positive, and you're asked to obtain the true prevalence of the disease.
Nonetheless, they can be worked in more or less the same basic ways. One avenue is to imagine a generic population, where $q$ is the proportion of the population that has TB; that's what we're going to solve for. From that portion, $0.99q$ tests positive. From the $1-q$ that doesn't have TB, $0.05(1-q)$ also tests positive. That makes $0.99q + 0.05(1-q) = 0.05+0.94q$ that tests positive overall. Since we're told that $5$ percent of those that test positive actually have TB, we must have
$$
0.99q = 0.05(0.05+0.94q)
$$
from which we can solve for $q$.
Another possibility is to use Bayes's Formula directly:
$$
0.05 = P(\text{has TB} \mid \text{tests positive})
= \frac{P(\text{tests positive} \mid \text{has TB})
P(\text{has TB})}
{P(\text{tests positive})}
$$
where
\begin{align}
P(\text{tests positive})
& = P(\text{tests positive} \mid \text{has TB})P(\text{has TB}) \\
& + P(\text{tests positive} \mid \text{doesn't have TB})
P(\text{doesn't have TB})
\end{align}
Filling in the values we have gives us
$$
0.05 = \frac{0.99q}{0.99q + 0.05(1-q)}
$$
and again we can solve for $q$. It's essentially the same equation, but we arrived at it more heuristically in one case, and more symbolically in the other.
I will answer the first questions, the second is very similiar.
$$p(\text{covid}|\text{positive}) = \frac{p(\text{positive}|\text{covid})p(\text{covid})}{p(\text{positive}|\text{covid})p(\text{covid})+ p(\text{positive}|\text{not covid})p(\text{not covid})} =\\
\frac{0.985\times0.33}{0.985\times0.33 + 0.2\times0.66} \sim 70\%
$$
This is why doctors make patients take more tests if a test is known to give false positives.
See pic for a geometric interpretation for this kind of problems: think about plugging a number of random event at top.
Bye
Best Answer
Legend. $$P(\text{actually sick} \mid \text{all $n$ results are positive})\\ =\frac{P(\text{actually sick and all $n$ results are positive})}{P(\text{all $n$ results are positive})}\\ =\frac{pv^n}{pv^n+(1-p)(1-f)^n}. $$ (The denominator corresponds to Branches 1 and 3 in the above diagram.)
P.S. That boldfaced assumption is not in general true: for example, the second of two successive COVID swab tests on the same subject is more probable than the first to register negative, due a lower concentration of antigens in the second reagent tube.