If you are struggling with the first problem, don't give up! Perhaps the "easiest" one is the last question. Depending on your perspective, the hardest problem is probably proving the first statement using the contrapositive. Let me give some hints about this one.

First, let's recall the statement that we are trying to prove:

If $x$ and $y$ are odd integers, then $xy$ is odd.

To form the contrapositive, we need to negate the hypothesis and conclusion and reverse the order of the implication. The negation of "$xy$ is odd" is "$xy$ is even." However, the negation of "$x$ and $y$ are odd integers" is a bit trickier. Before negating, notice that the original hypothesis is equivalent to "$x$ is an odd integer **and** $y$ is an odd integer." The negation of this statement is "$x$ is an even integer **or** $y$ is an even integer". Notice that the "and" in the original statement becomes an "or". The "or" in this case is an inclusive "or", which means one or the other or both. This is an application of a rule of logic called DeMorgan's Law, which you can read more about here. One can quickly write down a truth table to convince themselves this must be the case. In addition, it matches our use of everyday language. Think of some statements that involve "and" and ask yourself what it would mean if the statement was false. (There are other cases to consider, but this should be enough to convince that our model is accurate.) OK, in summary, the contrapositive is:

If $xy$ is even, then $x$ is an even integer or $y$ is an even integer.

To prove the contrapositive directly, you would start your proof with something like: "Assume that $x$ and $y$ are integers such that $xy$ is even." Your goal is to conclude that $x$ is an even integer or $y$ is an even integer.

If you wanted to finish this up by using a proof by contradiction, there wouldn't be anything wrong with that. In this case, you would write something like: "For sake of a contradiction, assume that it is not the case that either $x$ is an even integer or $y$ is an even integer." This implies that both $x$ and $y$ are odd. Now, do what we have done in lots of problems. What can you conclude?

Another approach could be to first prove as a lemma that if $p$ is prime and divides $xy$ where $x,y\in\mathbb{Z}$, then $p$ must divide either $x$ or $y$. A special case of this is when $p=2$. By the way, if you remove the requirement that the divisor be prime, it is no longer true. Y'all provided a counterexample to this already on a previous homework assignment.

If people ask, I'll provide more hints on the other problems (which are not nearly as difficult as the one discussed above).