In Algebra by Serge Lang, the author asserts, to prove the existence of a field extension where an irreducible polynom has a root, that if you take one set $A$ and a cardinal $\mathcal{C}$, that you can find a set $B$ such that $\text{card}(B)=\mathcal{C}$ and such that $A \cap B=\emptyset$.
How can one build $B$ ?
On
If you use the axiom of foundation, you can set $$ B := \left\{\{A,c\} \,\big|\, c \in \mathcal{C} \right\} \text{.} $$ $B$ obviously has cardinality $\mathcal{C}$, and if $x \in B \cap A$, then the axiom of foundation is violated because you then have an $\in$-circle $$ x \in A \in \{A,c\} = x \text{.} $$
Note that this assumes that cardinalities are sets - in particular, that $\mathcal{C}$ is a set. If your definition of cardinal doesn't make them sets, e.g. if your idea of a cardinal is the class of all sets with a given cardinality, then instead of $\mathcal{C}$, pick some set $C$ with $card(C) = \mathcal{C}$.
On
There's an ordinal $\alpha$ such that all ordinals $\beta\ge\alpha$ are not in $A$. Take $\mathcal C$ ordinals starting at $\alpha$ as your set $B$.
Why does such ordinal $\alpha$ exist? Assume it doesn't, then the set $A$ contains arbitrarily big ordinals. By a transfinite recursion, we can construct a set of increasing ordinals from $A$ indexed by all ordinals. Just take an ordinal bigger than the supremum of all ordinals so far. This is a contradiction since ordinals are too numerous to form a set.
On
Given any two sets, $A$ and $B$ we can find $A'$ and $B'$ such that $A'\cap B'=\varnothing$ and $|A|=|A'|$ and $|B|=|B'|$.
Simply take $A'=\{0\}\times A$ and $B'=\{1\}\times B$.
Since $A$ is fixed, using Cantor's theorem we know that that there exists some $X\subseteq A$ such that there are no ordered pairs of the form $\langle X,a\rangle\in A$ (for any $a$, that is). So now we can take $B'=\{X\}\times B$.
Another approach, that avoids having to deal with ordinals or to invoke the axiom of foundation, is to note that there is a set $t$ not in $\bigcup\bigcup A$ (just on cardinality grounds: for any set $X$ there is a set $z$ not in $X$). Therefore if $B$ has size $\mathcal C$, letting $B'=\{t\}\times B$, we see that $B'$ is both of size $\mathcal C$, and disjoint from $A$. The point is that the usual set theoretic formalization of ordered pairs gives us that $(t,a)=\{\{t\},\{t,a\}\}$, so if such a pair $(t,a)$ is in $A$, then $t$ is in $\bigcup\bigcup A$.