Assume that $n \ge 1$.
Let $\alpha_n$ denote the least countable ordinal that is not $\Delta_n^L$-definable.
Let $x_n$ denote a real which encodes all countable ordinals that are $\Delta_n^L$-definable.
Let $y_n$ denote a real which encodes all subsets of $\omega$ that are $\Delta_n^L$-definable.
Let $\beta_n^z$ denote the least countable ordinal that is not $\Delta_n$-definable in $L[z]$ with a predicate for $z$.
Question $1$: Which ordinal is larger, $\alpha_{n+1}$ or $\beta_n^{x_n}$? If they are equal, why?
Question $2$: Which ordinal is larger, $\alpha_{n+1}$ or $\beta_n^{y_n}$? If they are equal, why?
I am interested in all cases ($V = L$; $V \neq L$; no sharps exist; sharps exist for all subsets of $\omega$; sharps exist, but not for all subsets of $\omega$).
As @HanulJeon points out, it is very important how exactly $x_n$ is defined. But let's suppose we define it in the following reasonably minimal manner: $x_n$ is the $L$-least wellorder of $\omega$ in ordertype $\alpha_n$, and $y_n$ is the $L$-least real which is Turing above all subsets of $\omega$ which are $\Delta_n^L$-definable.
I'm also going to assume $n>0$, as I'm not exactly sure how to interpret $\Delta_0^L$-definability.
Then $\alpha_{n+1}>\beta_n^{x_n}=\beta_n^{y_n}$.
The main point here is that $\alpha_n<\alpha_{n+1}$. For this, first suppose $n=1$. Let $\lambda_1$ be the least ordinal such that $L_{\lambda_1}\preccurlyeq_1 L$. Then $\{\lambda_1\}$ is $\Sigma_2^L$-definable. But using the parameter $\lambda_1$ it is easy to compute $\alpha_1$ itself, so $\alpha_1<\alpha_2$. When $n>1$ it is similar. (Actually $\lambda_1=\alpha_1<\omega_1^L$, but that is special to the case $n=1$; easily $\omega_1^L<\lambda_2$.)
Similarly, for each real $x$, we have $\beta_n^x<\beta_{n+1}^x$.
Now because $\alpha_n<\alpha_{n+1}$, we get that $x_n$ and $y_n$ are both $\Delta_{n+1}^L$-definable (this also uses that from any ordinal $\beta<\omega_1^L$, we can $\Sigma_1^L$-in-parameter-$\beta$ compute the least $\gamma\geq\beta$ such that $L_\gamma$ is pointwise definable, and applying this to $\beta=\alpha_n$, we can determine $x_n$ and $y_n$ from the theory of this $L_\gamma$).
It follows that $\beta_n^{x_n}<\beta_{n+1}^{x_n}=\alpha_{n+1}$ and $\beta_n^{y_n}<\beta_{n+1}^{y_n}=\alpha_{n+1}$.