For question $2a$.) i so far put down:
The recursive definition of $$S(n+1) = S(n) + \left[\left(\frac{1}{2^n}\right) - 1\right]$$
$2b$.) Induction hypothesis, $n\ge0$, $S(n) = 1 - \left(\frac{1}{2^n}\right)$
Induction step, $S(n+1) = 1-\left(\frac{1}{2^{n+1}}\right)$
$$S(n+1) = S(n) + \left(\frac{1}{2^{n+1}}\right) = 1-\left(\frac{1}{2^{n}}\right)+\left(\frac{1}{2^{n+1}}\right)$$
Your recursive definition is not correct in two ways. Think about subtracting $S(n+1)-S(n)$. Most of the terms cancel. The right side should be the one that does not, and neither of your two is it.
Then in your induction step, you have the hypothesis correct. You should then add $S(n+1)-S(n)$ to that so you have $S(n+1)$ on the left. You need to make the right side match your closed form, so you need to get to $S(n+1)=1-1/(2^{n+1})$, which you have not done. Your line Induction step assumes what you are trying to prove.