To answer the first three questions negatively, the key is to show that measurable weakly Shelah cardinals are limits of *weakly* Shelah cardinals.

To see this, suppose that $\kappa$ is weakly Shelah and measurable. Let $j : V\to M$ be an elementary embedding with critical point $\kappa$. We claim that $\kappa$ is weakly Shelah in $M$. We will show that $\kappa$ has the weak Shelah property with respect to any increasing function $f : \kappa\to \kappa$. (It suffices to handle increasing functions.) In other words, we will find a cardinal $\nu < \kappa$ and an elementary embedding $i : M\to N$ definable over $M$ such that $\text{crit}(i) = \nu$, $i(\nu) > \kappa$, and $V_{i(f)(\kappa)}\cap M\subseteq N$. Since $\kappa$ is weakly Shelah in $V$, there is a cardinal $\nu < \kappa$ and a definable elementary embedding $\bar i : V\to \bar N$ such that $\text{crit}(\bar i) = \nu$, $\bar i(\nu) > \kappa$, and $V_{\bar i(f)(\kappa)}\subseteq \bar N$. Let $i = j(\bar i)$ and let $N = j(\bar N)$. Thus $i$ (resp. $N$) is defined over $M$ by the same formula as $\bar i$ (resp. $\bar N$) but with the parameters shifted via $j$. 

We now verify that $i$ is as desired. It's pretty easy to see $i(\nu) = \bar i(\nu) > \kappa$. Moreover by the elementarity of $j$, $V_{j(\bar i(f)(\kappa))}\cap M=  j(V_{\bar i(f)(\kappa)})\subseteq j(\bar N) = N$. To show $V_{i(f)(\kappa)}\cap M\subseteq N$, it therefore suffices to show $j(\bar i(f)(\kappa)) \geq i(f)(\kappa)$, which is where we use that $f$ is increasing: $$j(\bar i(f)(\kappa)) = j(\bar i)(j(f))(j(\kappa))\geq j(\bar i)(j(f))(\kappa) = i(j(f))(\kappa) = i(f)(\kappa)$$

This shows $\kappa$ is weakly Shelah in $M$. Hence by the standard reflection argument, $\kappa$ is a limit of weakly Shelah cardinals. We can conclude that the least weakly Shelah cardinal is not measurable, so in particular it is not Shelah.

As for consistency strength, it is not even completely obvious from this that the existence of a Shelah cardinal implies the consistency of a weakly Shelah cardinal, but this too is true. Proving this involves examining the witnessing ordinal of (weakly) Shelah cardinals. The witnessing ordinal of a (weakly) Shelah cardinal $\kappa$ is the least cardinal $\lambda$ such that $\kappa$ is witnessed to be (weakly) Shelah by extenders in $V_\lambda$. The main observation about $\lambda$ is that $\kappa < \text{cf}(\lambda) \leq 2^\kappa$. It follows pretty easily that the witnessing ordinal for the first weakly Shelah is below the second Shelah cardinal. (Incidentally, it seems one must go through this kind of calculation just to show that two Shelahs are stronger than one.)

Still thinking about the weakly hyper-Woodin question.