This seems a bit challenging for me, I tried looking for ways to Alice to identify 5 genuine coins anytime but could not do better than 4 for some cases. I think I have a proof, that Alice may not always find 5 coins: At any time, we can divide the 8 coins into three disjoint subsets: $A$: Coins that were used to pay measurement fee $B$: Coins that were not used as fee (are still available for Alice) and have been distinguished as genuine coins $C$: Coins that are still available for Alice but are not yet known whether genuine or counterfeit, and possibly the counterfeit coin (if it was not used as a fee for measurement). Alice wins if at any point $|B| \geq 5$. There are two ways to win: a) Identifying which coin is counterfeit (and having at least 5 genuine coins left) b) Not identifying which coin is counterfeit, just distinguishing 5 coins as genuine. The b) case would mean that if Alice unluckily happens to choose genuine coins for fee (possible since she doesn't know which one is counterfeit), then $|C| \geq 2$ at any time, since she the counterfeit coin is in set $C$ and at least one genuine coin has to be unidentified for Alice not to be able to identify the counterfeit coin, so after just using two coins, $|B| = 8-|A|-|C| \leq 8-2-2=4$, thus Alice has to identify 5 genuine coins after just 1 measurement, which is of course impossible. Therefore, case a) is the only possibility. But here, either Alice has to identify the counterfeit coin after just 2 measurements if she paid with two genuine coins, or after 3 measurements if she once paid with the counterfeit coin. We'll focus on the first case: this case is possible, as there is no way to tell which coin is the counterfeit one after one measurement paid with a genuine coin, and hence the second measurement might happen to be paid by a genuine coin. So we may assume that Alice paid the for the first two measurements with genuine coins. We can divide the available coins (union of $B$ and $C$) at each measurement into three subsets again: $X$: the left plate of the balance scale $Y$: the right plate $Z$: the remaining, unmeasured coins It only makes sense to measure with the same amount of coins on the plates, so we may assume that $|X| =|Y|$. Take the first measurement: 1st case: if $|X|=|Y| \leq 2$ then assume the case when the counterfeit coin is in $Z$. After this, Alice can identify 2 or 4 genuine coins, however, there will be three coins remaining (that were in $Z$) and one of which is the counterfeit coin, and after the second measurement, she can not tell which coin is the counterfeit, because she doesn't even know whether the counterfeit coin is lighter or heavier. 2nd case: if $|X| = |Y|=3$ then assume the case when the coin is in one of the plates. Since we do not know whether the counterfeit coin is heavier or lighter, we do not know which plate has it, and any next measurement cannot distinguish the counterfeit coin at any case. Thus we're done, Alice can not always achieve this.

I think there might be a shorter proof for the b) case, using the fact that the counterfeit coin can be both heavier or lighter. That probably means that we can never exactly know which plate has the counterfeit coin, we can only realize which coin is counterfeit by identifying all other coins, or using it and realizing that the given answer was incorrect.