You didn't finish your question "if the draws are made..."

Made what?

Are the draws with replacement or without?

With replacement, sometimes it is easier with these kinds of problems just to enumerate the choices.

If you draw a 1 (1/5) probability, you have a 3/5 (3,4,5) chance that the numbers will be two or more apart. So you multiply these two events to get the probability. The rest of the probabilities are similar.

1: (1/5) * (3/5)

2: (1/5) * (2/5)

3: (1/5) * (2/5)

4: (1/5) * (2/5)

5: (1/5) * (3/5)

Since each of these events are independent, you add them all together, giving 12/25 or a bit less than 1/2.

Without replacement, the same idea applies but now you remove one the card you drew, making it easier to get one two or more apart:

1: (1/5) * (3/4)

2: (1/5) * (2/4)

3: (1/5) * (2/4)

4: (1/5) * (2/4)

5: (1/5) * (3/4)

Giving 12/20 = 3/5, or a bit more than 1/2.