How do we prove that $L_E = \{ \langle A, B \rangle : L(A) = L(B) \}$ is neither regular nor context-free?
One possible way, is to use the pumping lemma, but one needs to be careful about the details. For simplicity let's discuss only the pumping lemma for regular languages (but I believe a similar method can be used when considering the CFL-pumping lemma).

Let's assume that $\langle A \rangle$ is a well-defined way to encode DFAs, that begins with describing the number of states, then list the transition for each state, etc.
Also assume that any string that is not "well formatted" under this encoding just describes a DFA that rejects all words.

Let $n$ be the pumping length.

Let $A=B$ be a $2^n$-state DFA that accepts all words. The encoding of $A$ begins with stating it has $2^n$ states, that is the first part is of length $\log(2^n)=n$ bits at least. 

Now let's use the pumping lemma on $w=\langle A,B\rangle$ (which is longer than $n$, so the pumping lemma applies). We can split $w$ into $xyz$ such that $|xy|\le n$, that is, the part we pump "lies" in the part that describes the amount of states in $A$.
Now, we pump the word down. That is, consider the words $w'=xz$ which should be also in $L_E$ by the pumping Lemma.

However, we can interpret $w'$ as being $w'=\langle A',B\rangle$ where $A'$ is a malformed encoding (it begins with stating the DFA has less then $2^n$ states, but then it describes transitions of $2^n$ states, which is inconsistent with the claimed number of states).

Therefore, by our assumption $L(A')=\emptyset$ while $L(B)=\Sigma^*$. Thus, $\langle A',B\rangle \notin L_E$ and we reached a contradiction.