Yes, functions can be defined using set-builder notation. But this is not the most common approach.
However, what you have written down is not really well-formed. Let me address this first; then we can end on a positive note.
Compare your expressions with their correct counterparts, so that you may avoid these mistakes in the future (some things are common abuses of notation, but I strongly suggest you get down to writing the tedious full form until you can recognise a well-formed expression within a second):
$$\begin{array}{c|c}
\forall x \in X, \exists! y \in Y|(x,y)\in f &\forall x \in X:\exists! y\in Y: (x,y) \in f\\
f= \{(x,y)|(x\in X)\land(\exists!y \in Y) \land ((x,y)\in f\} & f = \{(x,y) \mid \forall x \in X: \exists! y \in Y: (x,y)\in f\}
\end{array}$$
and a similar problem arises with the expression for $F$.
So, what is usually done is the following. Let $\rm fn$ be the unary predicate given by:
$$\mathrm{fn}(f) = \forall z \in f: (\exists x,y: z = (x,y)) \land (\forall z' \in f: x_z = x_{z'} \to z = z')$$
That is: "$f$ is a collection of ordered pairs $z = (x_z, y_z)$, and each first coordinate $x_z$ is unique in $f$."
Note that we haven't specified the domain or codomain of $f$ yet. We can use the following binary predicates:
\begin{align}
\mathrm{dom}(f,X) &= \mathrm{fn}(f) \land \forall x: (x \in X\leftrightarrow \exists y: (x,y) \in f)\\
\mathrm{cod}(f,Y) &= \mathrm{fn}(f) \land \forall y: ((\exists x: (x,y) \in f) \to y \in Y)
\end{align}
Then we can define the set of functions from $X$ to $Y$, $Y^X$, by:
$$Y^X = \{f \subseteq X \times Y\mid \mathrm{fn}(f) \land \mathrm{dom}(f,X) \land \mathrm{cod}(f,Y)\}$$
where it is not hard to show that $f \subseteq X \times Y$ makes the $\rm cod$ condition a consequence of the other two.
All being formulated in the first-order language of set theory, this is a first-order definition. The tell-tale sign is that we have no need for quantifying over predicates, etc..
You're on the right track, but you're having trouble with the endpoints.
The points $(-3, -2)$ and $(4, -2)$ are on the graph in a), which means that your domain should actually be $x \in [-3, 4]$ in interval notation to indicate that $x = -3$ and $x = 4$ are included. This would be reflected in the set-builder notation by using inequalities with "or equal to", so that you'd have $\{x \in \Bbb R \mid -3 \le x \le 4\}$.
For part c), you're again almost right, except for the endpoints. Their $x$- and $y$-coordinates need to be included, so you'd need brackets for interval notation, and all $\le$'s for the set-builder notation. So, for the range, I would write $\{y \in \Bbb R \mid -3 \le y \le 3\}$, using $y$ rather than $x$.
Now, in addition to filled-in circles, some graphs have arrows. These indicate that the graph "keeps going" in (roughly) whatever direction the arrows point. In b), this would be reflected as an interval $x \in (-\infty, 3]$ for the domain.
Notice for b) that $x$ needs to only be "at most $3$" (not "at least" anything), and thus you'll only need a single inequality, rather than the compound ones you would use on graphs that have a definite starting and ending point.
I'll let you give the rest a shot; most of what you had was spot-on.
Best Answer
You want to find the domain of $\displaystyle \frac{1}{\sqrt{x^2-4}}$.
We need two things: the denominator must be defined, and nonzero.
The denominator is equal to zero when $x = \pm 2$.
We also have $x^2-4<0$ if $-2<x<2$, so the denominator is not defined for $x \in (-2,2)$. *
Therefore, the function is NOT defined on $x \in [-2,2]$, and thus the function IS defined on $\boxed{\{x \in \mathbb{R}|x<-2~\text{or}~x>2\}}$
Or more simply, $\boxed{\{x \in \mathbb{R}\mid|x|>2\}}$
*Notice that $x^2-4<0 \implies x^2<4 \implies |x|<2$, since $\sqrt{x^2}=|x|$.
Then, since $|x|=2$ also makes $f(x)$ undefined, we have f(x) is defined for reals where $|x|>2$.