.
In this my post, I will not count the number of correct and wrong statements in the problem.
I will concentrate/focus entirely on statement III, ONLY.
Integral with p > 0 does exist (= converges) if and only if 0 < p < 1.
For the REFERENCE, see, for example,
https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_ImproperIntegrals.html
or your textbook/handbook (which, as I assume, is ALWAYS with you).
If the presumption in III is correct with p > 0, it means that 0 < p < 1.
Then, if q > p, there are two possibilities:
(a) q is still less than 1: q < 1, and then
integral does exist (= converges).
(b) q >= 1, and then integral does NOT exist (= does NOT converge).
+-------------------------------------------------------------+
| THEREFORE, the statement III, as "if-then" statement, |
| considered for all possible cases, is FALSE. |
+-------------------------------------------------------------+
Sometimes (for some combinations of p and q), it is correct.
For other combinations, it is wrong.
+---------------------------------------------------------+
| But as a UNIVERSAL "if-then" statement, it is FALSE. |
+---------------------------------------------------------+
For example, the case p= 1/2, q= 2 is a COUNTER=EXAMPLE:
at p= 1/2, the presumption of III is correct, but the conclusion of III at q= 2 is WRONG.
Thus part (III) is solved and carefully/thoroughly explained.
-------------
Notice that interpretation given in the post by Edwin, does not work in case III.
It does not work, because the curve can be entirely below the curve
(as it happens at p= 1/2, q= 1/3, 0 < x < 1), when the integral does exists,
but it can be entirely above that curve (as it happens at p= 1/2, q= 2/3, 0 < x < 1), when the integral still exists,
or it can be entirely above that curve (as it happens at p= 1/2, q= 2, 0 < x < 1), when the integral does not exist.
Plots y = 1/x^(1/2) (red), y = 1/x^(1/3) (green), y = 1/x^(2/3) (blue), y = 1/x^2 (magenta).