a) Yes, the product $X\times_k Y$ of two varieties over an algebraically closed field $k$ is a variety. To prove it, reduce to affine varieties and for those use:
b) If the field $k$ is algebraically closed and if $A$ and $B$ are $k$-algebras without zero divisors, then their tensor product $A\otimes_kB$ also has no zero divisors.
(Whether $A$ or $B$ is finitely generated is irrelevant.)
This is proved in Iitaka's Algebraic Geometry, page 97 (Lemma 1.54)
Edit: And if $k$ is not algebraically closed...
... all is not lost!
Let $k$ be a field and $R$ be a $k$-algebra which is a domain and with fraction field $K$.
If the extension $k\to K$ is separable and if $k$ is algebrically closed in $K$, then for all $k$-algebras $S$ which are domains, the $k$-algebra $R\otimes_k S$ will be a domain.
(Beware that separable means universally reduced. If $k\to K$ is algebraic, separability coincides with the usual notion.)
With luck, one of $A$ or $B$ may play the role of $R$ and $A\otimes_k B$ will be a domain.
As an illustration, any intermediate ring $k\subset R\subset k(T_1,\cdots,T_n)$ (where the $T_i$'s are indeterminates) satisfies the conditions above and will remain a domain when tensorized with a domain.
Bibliography
For the product of algebraic varieties I recommend Chapter 4 of Milne's online notes.
For the tensor product of fields, you might look at Bourbaki's Algebra, Chapter V, ยง17.
New Edit
At Li's request I'll show that $X\times Y$ is irreducible.
Let $$X\times Y=F_1\cup F_2$$ with $F_i$ closed and consider the sets $X_i=\lbrace x\in X\mid \lbrace x \rbrace\times Y\subset F_i\rbrace$.
Since $$\lbrace x \rbrace\times Y=[(\lbrace x \rbrace\times Y)\cap F_1]\cup [(\lbrace x \rbrace\times Y)\cap F_2]$$ we see, by irreducibility of $\lbrace x \rbrace\times Y$ (isomorphic to $Y$), that each vertical fiber $\lbrace x \rbrace\times Y$ is completely included in (at least) one of the $F_i$'s. In other words, $$X=X_1 \cup X_2.$$ It suffices now to show that the $X_i$'s are closed, because by irreducibility of $X$ we will then have $X_1=X$ (say) and thus $X\times Y=F_1$: this will prove that indeed $X\times Y$ is irreducible.
Closedness of $X_i$ is proved as follows:
Choose a point $y_0\in Y$. The intersection $(X\times \lbrace y_0 \rbrace)\cap F_i$ is closed in $X\times \lbrace y_0 \rbrace$ and is sent to $X_i$ by the isomorphism $X\times \lbrace y_0 \rbrace \stackrel {\cong}{\to}X$. Hence $X_i\subset X$ is closed. qed.
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.
Best Answer
First notice that there exists a dense open subset $V$ of $Y$ such that $\beta^{-1}(V)\to V$ is an isomorphism. Above this $V$, $X\times_Y Y'$ is then canonically isomorphic to $f^{-1}(V)$.
Let $R$ be an algebraically closed field, $Y$ a curve with a node $y_0$ and $Y'\to Y$ the normalization map. Let $X$ be any normal variety with a surjective morphism to $Y$. Then $X\times_Y Y'\to X$ is finite (because $Y'\to Y$ is finite). If $X\times_Y Y'$ were irreducible, then $(X\times_Y Y')_{\mathrm{red}}$ would be integral, finite and birational (by (1) above) to $X$. As $X$ is normal, $(X\times_Y Y')_{\mathrm{red}}\to X$ would be an isomorphism. Let $x_0\in f^{-1}(y_0)$, then the pre-image of $x_0$ by this isomorphism is $\{x_0\}\times \beta^{-1}(y_0)$. But the latter has two points because $y_0$ is a node. Contradiction.
By (1), the pre-image by $X\times_Y Y'\to Y'$ of $\beta^{-1}(V)$ is irreducible and isomorphic to $f^{-1}(V)$. So there exists a unique irreducible component $X'$ in $X\times_Y Y'$ dominating $Y'$. It is equal to the Zariski closure of $f^{-1}(V)\times_V \beta^{-1}(V)$ in $X\times_Y Y'$.
$X'$ is birational to $X$ by (3).