How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like?

Yes, they're all empty sets. For example, $\emptyset \times A$ consists of all pairs of the form $(o,a)$ with $o \in \emptyset, a \in A$. But the empty set has no elements, hence $\emptyset \times A$ has no elements, hence $\emptyset \times A$ is the empty set. A similar argument works for the other two sets.

Here is how this problem can be interpreted in terms of cardinalities. For any sets $A,B$ the cardinality of $A \times B$ is the product of cardinalities of $A$ and $B$. Hence the cardinality of $\emptyset \times A$ is just $0 \cdot |A| = 0$ so $\emptyset \times A$ has $0$ elements, and hence $\emptyset \times A = \emptyset$. And a similar argument will work in the other two cases.

A slightly more formal rephrase of the other answer is to calculate which elements are in $A \times \emptyset$: for any $p$, \begin{align} & p \in A \times \emptyset \\ \equiv & \;\;\;\;\;\text{"definition of $\times$ on sets"} \\ & \textrm{isPair}(p) \;\land\; \textrm{fst}(p) \in A \;\land\; \textrm{snd}(p) \in \emptyset \\ \equiv & \;\;\;\;\;\text{"$x \in \emptyset \equiv \textrm{false}$ for any $x$; simplify"} \\ & \textrm{false} \\ \equiv & \;\;\;\;\;\text{"$x \in \emptyset \equiv \textrm{false}$ for any $x$"} \\ & p \in \emptyset \\ \end{align} and therefore, by set extensionality, $A \times \emptyset = \emptyset$. A very similar proof of course goes for $\emptyset \times B = \emptyset$, and obviously either of these implies $\emptyset \times \emptyset = \emptyset$.