are regular languages closed under division
If $L_1$ is regular and $L_2$ is any language, then $L_1 / L_2$ is regular. In fact, the construction of a DFA for $L_1 / L_2$ is almost trivial (with one caveat): It's the same as the DFA for $L_1$, except that the "final" states are the the states which can reach the original accepting states given a string from $L_2$.

Suppose that $M_1 = (Q, \Sigma, \delta, q_0, F_1)$ is the DFA which accepts $L_1$. Then a DFA which accepts $L_2$ is $M_2 = (Q, \Sigma, \delta, q_0, F_2)$, where:

$$F_2 = \left\{ q \in Q : \exists x \in L_2. \delta(q,x) \in F_1 \right\}$$

Now here's the caveat: this proof sketch is not constructive! To turn this into an algorithm requires some way to enumerate or recognise strings in $L_2$, and remember, $L_2$ can be any language.

If $L_1$ is context free and $L_2$ is regular, $L_1 / L_2$ is always context-free (proof left as an exercise), but it is not necessarily regular. (To see why, consider $L_2 = \{ \epsilon \}$.)

EDIT I've just noticed that the definition in the question is different from the usual definition of "right quotient". The usual definition is:

$$A/B = \left\{x\in \Sigma^{*} : \exists y \in B. xy \in A\right\}$$

The definition in the question is:

$$A\div B = \left\{x\in \Sigma^{*} : \forall y \in B. xy \in A\right\}$$

My answer us for the usual definition, because these are standard results. The answers are similar for the definition in the question, except that the construction is slightly different. (See David Smith's comment.)

I suspect there is a typo in the original question.