Here is an analysis of the first exercise you did. It's a good first attempt, but there are several errors:

I don't quite understand your first induction proof. The base case is correct. The inductive step should look like this:

Assuming that , we have .

Your first proof by contradiction looks incorrect.

Mistake 1: A proof by contradiction would say "there issomesuch that ". You don't get to pick the for which this happens.

Mistake 2: You look at , but is not even defined in this problem.

An actual proof by contradiction here would require using the well-ordering property of the integers to get a least counterexample. This is essentially the same as induction, but less elegant.

I'm not sure I understand your reasoning by analysis, but it seems like an informal argument, and not a mathematical proof.

The proof by induction is the best way to go here.

The second question looks almost correct. For a cleaner argument follow what I did above for the first exercise.

The last one is completely wrong. You seem to think that a set is a number. This is false. A set is a collection of objects. Here is a specific example:

The set A has 2 elements. The 4 subsets of are:

, , ,

A proof by induction for the last problem would involve taking one element of the set, let's call it , and counting the subsets that exclude , and the subsets that contain , then adding these results together.