If we have a number of beans ending in a 5 and we double it, we will get a number of beans ending in a 0 (since \(5\cdot 2 = 10\) ). Then if we subtract 5, we will once again get a number of beans ending in a 5. Thus if on any day we have a number ending in a 5, the next day will also have a number ending in a 5.
If you don't start with a number of beans ending in a 5 (on day 1), the above reasoning is still correct but not helpful. For example, if you start with a number ending in a 3, the next day you will have a number ending in a 1.
Part (b) is the base case and part (a) is the inductive case. If on day 1 we have a number ending in a 5 (by part (b)), then on day 2 we will also have a number ending in a 5 (by part (a)). Then by part (a) again, we will have a number ending in a 5 on day 3. By part (a) again, this means we will have a number ending in a 5 on day 4
The proof by induction would say that on every day we will have a number ending in a 5, and this works because we can start with the base case, then use the inductive case over and over until we get up to our desired \(n\text{.}\)