The following definition of a large cardinal property combines parts of the definitions of "Shelah cardinal" and "Woodin cardinal":

A cardinal $\kappa$ is *weakly Shelah* if for all $f : \kappa \to \kappa$ there is some $\alpha < \kappa$ that is closed under $f$ and there is some elementary embedding $j : V \to M$ (where $M$ is a transitive class) such that $\operatorname{crit}(j) = \alpha$ and $j(\alpha) > \kappa$ and $V_{j(f)(\kappa)} \subset M$.

Questions:

Is every weakly Shelah cardinal Shelah?

Is the least weakly Shelah cardinal Shelah?

Is every weakly Shelah cardinal measurable?

What is the consistency strength of ZFC + "there is a weakly Shelah cardinal"? In particular, how does it relate to weakly hyper-Woodin cardinals as defined by Schimmerling?

Here's what I know:

If $\kappa$ is a Shelah cardinal, then $\kappa$ is weakly Shelah. To see this, let $f : \kappa \to \kappa$. By the Shelah property applied to $f+1$, there is an elementary embedding $j : V \to M$ such that $\operatorname{crit}(j) = \kappa$ and $V_{j(f)(\kappa)+1} \subset M$. Because Shelah cardinals are Woodin, some cardinal $\alpha < \kappa$ is $\mathord{<}\kappa$-$f$-strong in $V$ and therefore $\mathord{<}j(\kappa)$-$j(f)$-strong in $M$ by elementarity. In particular $\alpha$ is $(j(f)(\kappa)+1)$-$j(f)$-strong in $M$, and therefore also in $V$ because $V_{j(f)(\kappa)+1} \subset M$. It's not hard to see that $\alpha$ witnesses the weakly Shelah property of $\kappa$ in $V$ with respect to $f$.

On the other hand, if $\kappa$ is weakly Shelah, then $\kappa$ is a Woodin limit of Woodin cardinals and there are Woodin cardinals above $\kappa$. Clearly the definition implies $\kappa$ is Woodin. Then by considering various $f$ we can obtain cofinally many $\alpha < \kappa$ as in the definition, and $j(\alpha) > \kappa$ implies $\alpha$ is a limit of Woodin cardinals, so $\kappa$ is a limit of Woodin cardinals. Now applying the definition to the function $f : \kappa \to \kappa$ where $f(\alpha)$ is the successor of the least Woodin cardinal above $\alpha$ shows that there is a Woodin cardinal above $\kappa$.