The outcome of the first rounding step depends on the starting temperature's remainder modulo $9$, so it's reasonable to write the initial temperature as $9q+r$ for some $(q,r)$ and see what happens for different values of $r$. For example, we have:
$$\frac59(9q+0 - 32) = 5q - \frac{160}{9} \leadsto 5q - 18$$
where I denote by $\leadsto$ the result of rounding. Then we can do the reverse calculation:
$$\frac95(5q-18) + 32 = 9q - 0.6 \leadsto 9q - 1.$$
So we see that when $r=0$, we don't get the same number back. (If this is enough of a hint, stop reading here and try to solve the problem on your own.)
We can do the same calculation for other values of $r$. For $r=1$, we have
$$(9q + 1) ^\circ F = \left(5q - \frac{155}{9}\right)^\circ C \leadsto (5q-17)^\circ C = (9q + 1.2)^\circ F \leadsto (9q+1)^\circ F$$
so we do get the original number back.
Given 3 hours to do 15 problems, it is not unreasonable to take the time to just continue on in this way for $r=2,3,4,5,6,7,8$. We conclude that $r=1,3,5,7,8$ give back the original number, and the rest don't.
Starting at a temperature between $32$ and $1000$ means starting at $9\cdot 3 + 5$ and ending at $9\cdot 111 + 1$. So:
- $r=1$ occurs with $q = 4,5,\dots, 111$, a total of $108$ times.
- $r=3$ occurs with $q = 4,5,\dots, 110$, a total of $107$ times.
- $r=5$ occurs with $q = 3,4,\dots, 110$, a total of $108$ times.
- $r=7$ occurs with $q = 3,4,\dots, 110$, a total of $108$ times.
- $r=8$ occurs with $q = 3,4,\dots, 110$, a total of $108$ times.
The final answer is therefore $539$.
