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.
If this is correctly quoted, it is terribly incorrect. It looks like a copy/paste error: the text $$\mbox{The truth value of this statement is 'True' . . . but the statement is False}$$ makes me suspect that in an earlier draft, this was two examples - one of the form "there exists an $x$" and one of the form "for all $x$."
Best Answer
The first statement says that any integer number has an opposite. This is true but you havr to prove it starting from some axiomatic definition of integers, deducing the existence of the element $-a=b$.
The second statement says that there is an integer number that is the opposite of all ather numbers. Tis is false, because if, for the same $a$ we have: $a+b=0$ and $a+c=0$ with $b \ne c$, then we have a contradiction: $$ c=c+0=c+a+b=(a+c)+b=b $$