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.
If you're just trying to introduce a term that stands in for an arbitrary element of $A$ that satisfies the condition, you'd just want to use something like "$a\in A$ where $f(a)=3$".
You definitely would not want to use set builder notation like the first example, as this would be the subset of $A$ of all those things that satisfy $f(a)=3$. And using $b$ in the second example just adds a second variable that you don't really use.
Best Answer
The domain of your function $f$ is simply $\mathbb{R}$. Your third set, $\{ x | x \in \mathbb{R} \}$, is just a convoluted way of writing it, so it would be technically correct (but simply writing $\mathbb R$ is better).
The first one, $\{ x \in \mathbb{R} \}$, is not really set-builder notation, because you are missing the “selection” part and a set of the form $\{ a_1, \dots, a_n \}$ without a vertical bar is usually understood as listing all the elements of the set. People will probably understand what you mean, but again, writing $\mathbb{R}$ alone is shorter and clearer.
The second option is not a set but an assertion about $x$, so it is not directly an answer to the question “What is the domain of $f$?” That might be okay though, for example writing “The function defined by $f(x) = 3x + 2$ where $x \in \mathbb{R}$” is absolutely acceptable. Other options include:$$f : \mathbb{R} \to \mathbb R, \quad f(x) = 3x + 2$$ or $$f : \mathbb R \to \mathbb R, \quad x \mapsto 3x + 2.$$ In both of these, the first $\mathbb R$ in $\mathbb R \to \mathbb R$ specifies the domain as $\mathbb R$ and second specifies the codomain (also $\mathbb R$ is this case).