Although your notation $\omega_1^{CK}(\text{Ord}^M)$ suggests that you have some sort of generalized computability in mind, I'll take the question as being about all the well-orders of $\text{Ord}^M$ that are parametrically first-order definable over $M$. Such well-orders are elements of the next admissible set $M^+$, so the height of $M^+$ is an upper bound for their order-types. It might well be the least upper bound, but I'm not sure about that.
In the following answer, by Foreman-woodin model, I mean the model constructed by them in the paper "The generalized continuum hypothesis can fail everywhere.
Ann. of Math. (2) 133 (1991), no. 1, 1–35. "
Questions 1 and 3 have positive answer: In Foreman-Woodin model for the total failure of GCH the following hold:
1) For all infinite cardinal $\kappa, 2^{\kappa}$ is weakly inaccessible, and hence a fixed point of the $\aleph-$function,
2) If $\kappa \leq \lambda< 2^{\kappa},$ then $ 2^{\lambda}= 2^{\kappa}.$
In this model for all infinite cardinals $\kappa, 2^{\kappa}=\aleph_{ 2^{\kappa}}$ in particular for all fixed points $\kappa$ of the $\aleph-$function, $2^{\aleph_\kappa}=\aleph_{ 2^{\kappa}}$. Also note that in this model for all infinite cardinals $\kappa,$
if we let $\lambda=2^{\aleph_\kappa},$ then $\lambda \geq 2^\kappa,$ and $2^{\aleph_\kappa}=\aleph_\lambda.$ So both of questions 1 and 3 have a positive answer.
For your question 2, $\delta$ can be arbitrary large: Start with GCH+there exists a supercompact cardinal $\kappa$+ there are infinitely many inaccessibles above it. Now let $\delta$ be any ordinal $<\kappa.$ Force with Foreman-Woodin construction above $\delta$ (in the sense that let the first point of the Radin club added during their forcing construction be above $\delta$). In their final model (which is $V_\kappa$ of some extension of the ground model) for all infinite cardinals $\lambda <\kappa, 2^\lambda \geq \lambda^{+\delta}$. So if $F$ is defined in the ground model by $F(\kappa)=\kappa^{+\delta},$ then the acceleration rank of $F$ is $\delta$ (using GCH), and in the finial model for all infinite cardinals $\kappa, 2^\kappa \geq F(\kappa).$
Remark. I may note that we can not define the function $F$ in the ground model, such that it is the realization of power function in the extension, but we can find some inner model of the final extension in which $GCH$ holds and such a function $F$ is definable.
Best Answer
No. An early nontrivial constraint on the $\beth$ function comes from Kőnig's Theorem, that for all infinite $\kappa$, $\mathrm{cf}(2^\kappa)>\kappa$. This implies that we cannot have $\beth_\alpha = \aleph_{f(\alpha)}$ for all $\alpha$, when $f(1) = \omega$, nor when $f(\omega+1)$ is a cardinal below $\aleph_\omega$.
Another constraint is Silver's Theorem, that if GCH holds below a singular of uncountable cofinality, then it holds at that singular as well.
Other constraints come from Shelah's PCF theory. Shelah showed that if $\aleph_\omega$ is a strong limit, then $2^{\aleph_\omega} < \aleph_{\omega_4}$.