= Solution
Fix $i\in\{0,1\}$ and write $\kappa=\kappa_i$, $j=j_i$, and $M=M_i$. Every ordinal below $j(\kappa)$ is represented in the <ultrapower> by a function $\kappa\to\kappa$. Consequently, in $V_\lambda$,
$$
|j(\kappa)|\leq\kappa^\kappa=2^\kappa=\kappa^+,
$$
where the last equality uses the <Generalized continuum hypothesis>.
By <elementary embedding>[elementarity], $M$ regards $j(\kappa)$ as measurable and hence as a <strong limit cardinal>. Moreover $V_{\kappa+1}\subseteq M$, so $M$ and $V_\lambda$ have the same subsets of $\kappa$ and the same $\kappa^+$. It follows inside $M$ that
$$
\kappa^+=2^\kappa<j(\kappa).
$$
Thus, in the ambient $V_\lambda$, the ordinal $j(\kappa)$ is strictly larger than $\kappa^+$ but has cardinality at most $\kappa^+$. It cannot be a <cardinal number>. Applying this argument to both $i=0$ and $i=1$ proves that neither $j_0(\kappa_0)$ nor $j_1(\kappa_1)$ is a cardinal in $V_\lambda$, exactly as in <moved critical point is not an ambient cardinal under GCH>.
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