The method I tried:
5 x 4 x 3 = 60 different numbers
I solved it this way because if a number needs to be formed with a certain number of digits (with no restrictions - which I think 'as often as desired' means), you assign a scale to the given constituent digits and then pick the same number of digits as the new number needs, going from highest to lowest. Then multiply these digits by each other to give the amount of different numbers that could be created.
However my answer of 60 possible numbers conflicts with the textbook's answer of 64.
The next part of the question was finding how many 3-digit numbers can be formed using 2, 3, 4 and 5 using at most one each. I was able to get this question, by changing 2, 3, 4 and 5 to 1, 2, 3 and 4; then multiplying 4 by 3 by 2 to give 24 possibilities.
In each of the $3$ digit positions there are $4$ possibilities, so the number of possible combinations (reusing digits) is $4\cdot4\cdot4=4^3$.
If you cannot reuse digits, then the number of possibilities for the first digit is $4$; the number of possibilities for the second digit after choosing the first digit is $3$ since you've already used one; the number of possibilities for the third digit would be $2$. This gives you $4 \cdot 3 \cdot 2$.