For any language $L$, let $L_{\frac12-}$ be the set of first halves of the words in $L$ and $L_{-\frac12}$ be the set of second halves of the words in $L$. Also let $L^R$ be the set of reversals of words in $L$.
If $L$ is regular then so is $L^R$. Since $L_{-\frac12} = ((L^R)_{\frac12-})^R$ it is sufficient to show that if $L$ is regular then so is $L_{\frac12-}$.
Let $M = (Q,\Sigma,\delta,q_0,F)$ be a DFA such that $L(M) = L$.
Construct a DFA $N$ as follows:
- The set of states of $N$ is $Q \times P(Q)$ where $P(Q)$ is the power set of $Q$.
- The start state of $N$ is $(q_0,F)$.
- The set of accept states of $N$ is $\{(q,S) \in Q \times P(Q) \mid q \in S\}.$
- Define a function $\gamma : P(Q) \to P(Q)$ by $$\gamma(S) = \{q \in Q \mid {\rm there\ exists\ } a \in \Sigma {\rm\ such\ that\ } \delta(q,a) \in S\}.$$
The transition function $\bar\delta$ of $N$ is defined by $\bar\delta((q,S),a) = (\delta(q,a), \gamma(S)).$
Then $L(N) = L_{\frac12-}$.
Proof
Suppose $(q_0,S_0),\ldots,(q_n,S_n)$ is the path taken by $N$ on input $w = w_1\cdots w_n$, and $(q_n,S_n)$ is an accept state. Then $q_0,\ldots,q_n$ is the path taken by $M$ on input $w$ and $q_n \in S_n$.
From the definition of $\bar\delta$, there exist $q_{n+1}, \ldots q_{2n} \in Q$ and $v_1,\ldots,v_n \in \Sigma$ such that $q_{n+i} = \delta(q_{n+i-1}, v_i) \in S_{n-i}$, so $q_0,\ldots,q_n,q_{n+1},\ldots q_{2n}$ is the path taken by $M$ on input $w_1\cdots w_nv_1 \cdots v_n$. Also $q_{2n} \in S_0 = F$ so $w$ is the first half of a word accepted by $M$.
Conversely suppose $wv \in L(M)$ and $|w|=|v|=n$. Let $q_0,\ldots,q_{2n}$ be the path taken by $M$ on input $wv$, and let $(q_0,S_0), \ldots, (q_n,S_n)$ be the path taken by $N$ on input $w$. Then $q_{2n} \in F = S_0$ and so it follows by induction on $i$ that $q_{2n-i} \in S_i$ for $0 \leq i \leq n$. In particular, $q_n \in S_n$ so $N$ accpets $w$.
HINT: Start with a DFA $M=\langle Q,\Sigma,\delta,q_0,F\rangle$ for $A$, where the alphabet $\Sigma$ is the union of the alphabets for $A$ and $B$. Let $\bar\delta:Q\times\Sigma^*\to Q$ be the extended transition function: $\bar\delta(q,w)=q'$ if $M$ goes to state $q'$ when it starts in state $q$ and reads the word $w$.
To get a DFA for $A/B$ you need only modify $F$, the set of acceptor (final) states. Change it to
$$F'=\left\{q\in Q:\exists x\in B\,\left(\bar\delta(q,x)\in F\right)\right\}$$
I’ll leave it to you to show that this works.
Best Answer
Let $M_0$ and $M_1$ be two copies of a DFA for $L$, and let $M$ be their disjoint union. The initial state of $M_0$ will be the initial state of $M$. Replace each transition $p\overset{x}\longrightarrow q$ of $M_0$ by an $\epsilon$-transition $p\overset{\epsilon}\longrightarrow q'$, where $q'$ is the copy of $q$ in $M_1$. Replace each transition $p\overset{x}\longrightarrow q$ of $M_1$ by $p\overset{x}\longrightarrow q'$, where $q'$ is the copy of $q$ in $M_0$. Keep the acceptor states of $M_0$ and $M_1$ unchanged. You now have an NFA that accepts $\operatorname{even}(L)$.