There are already a couple of answers, but I think this question is a bit better than just a straightforward calculation problem, so here's one more.
You can reword the question into 5 separate questions like this:
- Here is a card of which you know nothing, except that it was the smallest in a random sample of 5 cards. Do you wish to exchange it for a card of which you do not have this information?
- Here is a card of which you know nothing, except that it was the second smallest in a random sample of 5 cards. Do you wish to exchange it for a card of which you do not have this information?
- And so on.
Casually slathering some Bayes onto the given information, then, we get that the cards that were smaller than average in the random sample are more likely to be smaller than average in actuality too, and vice versa.
Since we expect the cards that we know nothing about to be "average" by definition, we should exchange two of the smallest cards for certain, and the two biggest cards never.
The middle card in our hand is a bit trickier. Luckily, the problem has many symmetries, so we can reason that we got exactly the same information about the card being small and the card being big (there are two smaller cards, and two larger cards), so whatever the likelihood distribution, it is going to be symmetrical, and therefore the expected size of the third card should still be "average".
So exchanging the third card would only affect our hand's likely deviation from the point value average, and not the expected point value of the hand itself.
In conclusion, to maximize our hand's point value,
we should exchange either 2 or 3 cards.
Interestingly, we never used the total number of cards in this reasoning, so this strategy should maximise our points no matter how many cards in the deck. (As long as the cards are numbered with unique, consecutive integers, or in another suitably symmetrical manner.)