I’m afraid that you got them exactly backwards. Let’s take a look.
(a) For each integer $a$ there is an integer $b$ such that $a+b$ is odd.
Think of this in terms of a game. I give you an integer $a$, and you win if you can find an integer $b$ such that $a+b$ is odd; if you cannot find such a $b$, I win. The assertion (a) says that you can always win, no matter how I choose my $a$.
Suppose I give you the integer $101$; can you find an integer to add to it to make an odd integer? Sure: it’s already odd, so just add $0$. Suppose instead that I give you the integer $100$; can you find an integer to add to it to make an odd integer? $0$ won’t work, but $1$ will: $100+1=101$, which is certainly odd. Was there anything very special here about $101$ and $100$? No: all I used was the fact that $101$ is odd, so that adding $0$ was bound to give me an odd total, and the fact that $100$ is even, so that adding $1$ would give me an odd total. If I give you any odd integer $a$, you can use $b=0$ and be sure that $a+b=a+0=a$ will be odd. And if I give you any even integer $a$, you can use $b=1$ and be sure that $a+b=a+1$ is odd. Every integer is either even or odd, so no matter what integer $a$ I give you, you’re covered: you know how to pick a $b$ such that $a+b$ is odd. In other words, (a) is true: you do have a winning strategy.
(Of course there are other choices that work besides the ones that I’ve mentioned; mine are just the simplest.)
(b) There is an integer $b$ such that for all integers $a$, $a+b$ is odd.
Again you can think of this in terms of a game, with you picking $b$ and me picking $a$. The difference is that in this game you have to play first: you pick some integer $b$, then I pick my $a$, knowing what your $b$ is. You win if $a+b$ is odd, you lose if it isn’t, and (b) says that you have a winning strategy: there is some $b$ that you can pick that will make $a+b$ odd no matter what integer I pick for $a$.
But that’s clearly not true: if your $b$ is even, I’ll just let $a=0$, so that $a+b=b$ is even, and you lose. And if your $b$ is odd, I’ll pick $a=1$, so that $a+b=1+b$ is even, and again you lose. No matter how you play — no matter what integer you choose for your $b$ — I can beat you by choosing $a$ to make $a+b$ even.
You are correct for the first three, but not the fourth. To say that
$$A \subseteq B$$
means that every element of $A$ is also an element of $B$. The only element of $\{x\}$ is $x$ itself, and the only element of $B$ is $\{x\}$. These aren't the same, so the statement
$$\{x\} \subseteq \{\{x\}\}$$ is false.
In short, $x \notin \{\{x\}\}$, since the only thing in $\{\{x\}\}$ is $\{x\}$.
Best Answer
Bezout's identity. If $a$ and $b$ are relatively prime then there exist integers $x,y$ such that $ax+by=1$
Multiply both sides by $c$ for the first part and see what you can deduce.