[This answer has been edited to discuss the general case.]
I will assume that variety means irreducible (otherwise you could work on individual irreducible components). Then $f:X \to Y$ is dominant by assumption, and has closed image since its source is projective, thus it is surjective.
Thus $f^{-1}(Z) \to Z$ is also surjective.
Now, as described in e.g. this MO answer (or in a Hartshorne exercise, maybe in Section 4 of Chapter II), for the proper map $f$, the function $y \mapsto \dim f^{-1}(y)$ is upper semicontinuous, so if $z \in Z$, then the dimensions of $f^{-1}(z)$ are at least $\dim X - \dim Y$. This implies that $f^{-1}(Z)$ contains components of
codimension $1$. (The intuition is just that we can add the dimension of $Z$ and of a typical fibre. One way to make this precise is to note that if every component of $f^{-1}(Z)$ were of codimension at least $2$, then since it dominates $Z$, we would see that
a generic fibre would be of dimension $\dim X - \dim Y -1$, whereas we already noted that every fibre has dimension at least $\dim X - \dim Y$.)
In general you can't do better than this, because there are morphisms of $3$-folds (just to take an example) which are birational, but in which the preimage of some particular point $y \in Y$ is a curve. Then if you take $Z$ to be a generic codim'n one subvariety passing through $y$, its preimage will be the union of a codim'n subvariety of $X$ (the proper transform of $Z$) and the curve $f^{-1}(P)$. So in general you can't expect $f^{-1}(Z)$ to be equidimensional.
Note also that $f^{-1}(Z)$ can also contain multiple components of codimension $1$. (E.g. let $X \to Y$ be the blow up of a surface at a point, and let $Z$ be a curve that passes through the blown up point.)
Rereading the question, I see that the point of the question might be the equidimensionality, and so you might be interested in the counterexample involving $3$-folds. This MO question and answers gives one such example.
Best Answer
Question 1. No. Consider $f$ mapping $\Bbb A^1_k$ to $\operatorname{Spec} k[x,y]/(xy)$ by inclusion into the $x$-axis. Then $f^{-1}(\{y-\mathrm{axis}\})$ is a point which does not contain an irreducible component of $\Bbb A^1_k$.
Question 2. Note that any irreducible component is closed. Since $f$ is continuous, $f^{-1}(Y')$ is closed. We need to find conditions that ensure there's an irreducible component $X'$ of $X$ such that that $f^{-1}(Y')\cap X'$ isn't a proper subvariety of $X'$. Anything implying this is good enough. $Y$ irreducible or $f$ locally dominant works, for instance, but isn't necessary, since projection to a point inside a positive-dimensional variety satisfies the conditions as well.
Question 3. This is false even if you assume the statement in problem 1. Consider the map which takes any variety to a point inside $\Bbb A^1$. This is a counterexample.