Let me answer 4.8c first, and can fill in other hints/ideas when I have time.

For 4.8c, I'm a little confused by your method. Let me start backwards with your x's and y's and explain an idea from there, which hopefully helps.

It seems like at the end you're using x's for numbers chosen, and y's for numbers skipped. So your example xyyyyyxyxyyx should give the numbers 1, 7, 9, and 12, right? That's how I'm making sense of it at least.

In this way (ignoring the separators for a moment), any arrangement of $k$ x's and $n-k$ y's gives us $k$ numbers chosen from 1 to n. So how do we make sure each number is greater than $r$ apart? Note in terms of our arrangements, this means that each x needs to be separated by $r$ y's. Hence we set aside $(k-1)\cdot r$ (since there are $k-1$ spaces) y's to place between each pair of x's. Thus, we are left to arrange $k$ x's and $(n-k) - (k-1)\cdot r$ y's. This can be done in$$\binom{k+(n-k) - (k-1)\cdot r}{k} = \binom{n - (k-1)\cdot r}{k}$$ways.

Note: Since the x's and y's are arranged in any way at the end, it is possible for y's to occur BEFORE the first x. This means that the first value of r DOES NOT really have the restriction you were asking about.

Note 2: This idea is pretty much the same idea that is going on with the bijection described in the textbook. Try to compare them to see if it makes sense.

Hope this helps!