I claim that the answer to both questions is **yes**
.
Suppose that $A$ is a finite set, $j_{a}\in\mathcal{E}_{\lambda}$ for each $a\in A$ and $\gamma<\lambda$ is a limit ordinal. Then I claim that $\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is finite and the sequence $\{\textrm{crit}(j)|j\in\langle\{j_{a}|a\in A\}\rangle\}$ has order type $\omega$.

Let $M$ be the set of all strings $a_{1}...a_{n}$ in $A^{+}$ so that either $n=1$ or if $m<n$, then $\textrm{crit}(j_{a_{1}}[j_{a_{2}}]...[j_{a_{m}}])<\gamma$. Then $M$ is finite; if $M$ were infinite, then the tree $M$ would have an infinite branch consisting of all substrings of $(a_{n})_{n=1}^{\infty}$, but for such an infinite branch, the sequence of elementary embeddings $(j_{a_{1}}[j_{a_{2}}]...[j_{a_{n}}])_{n=1}^{\infty}$ contradicts the Laver-Steel theorem.

Let $F$ be the set of all maximal strings in $M$ (in other words, $x\in F$ if and only if $xa\not\in F$ whenever $a\in A$) and let $L=M\setminus F$. Then define a binary operation $*$ on $M$ according to the following rules:

i. $x*y=y$ whenever $x\in F$

ii. $x*a=xa$ whenever $x\in L,a\in A$

iii. $x*y=(x*z)*xa$ whenever $y=za$ and $x\in L$.

One can show that $*$ produces a well-defined total operation on $M$ the same way that one can show without resorting to large cardinals that the Laver table operation operation on $\{1,...,2^{n}\}$ is well-defined. Indeed, the structure $(M,*)$ is a generalization of the notion of a Laver table (the structure $(M,*)$ is self-distributive although we shall not use the self-distributivity of $(M,*)$ in this answer).

Define a mapping $\Phi:(M,*)\rightarrow(\mathcal{E}_{\lambda}/\equiv^{\gamma})$ by letting
$\Phi(a_{1}...a_{n})=j_{a_{1}}[j_{a_{2}}]...[j_{a_{n}}]$ whenever $a_{1}...a_{n}\in M$. Then one can show that $\Phi(x)*\Phi(y)=\Phi(x*y)$ by a double induction which is descending on the length of $x$ and for each individual $x$ one uses an induction which is ascending on the length of $y$. In other words, $\Phi$ is a homomorphism. However, we have $\Phi[M]=\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ since the algebra $M$ is generated by $A$. Therefore the algebra $\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is a finite algebra.

Since each algebra $\langle\{j_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is finite, the set $\{\textrm{crit}(j)|j\in\langle\{j_{a}|a\in A\}\rangle\}\cap\gamma$ is finite. Therefore, the set
$\{\textrm{crit}(j)|j\in\langle\{j_{a}|a\in A\}\rangle\}$ has order type $\omega$.