I suspect you didn't mean literally what you asked, but if you did, then the answer is yes. You required only that $X$ be a subset of $\mathcal A$, so even if $X$ consists entirely of finitary sentences and $\Delta$ is all of $X$, it wouldn't follow that it's an element of $\mathcal A$.

If you require $X$ to be an element of $\mathcal A$, to avoid the situation in the preceding paragraph, the answer is still yes, since you require $\Delta$ just to be some subset of $X$ consisting of finitary formulas. So, for example, $X$ could be the set of all finitary formulas of $L$, in which case $\mathcal A$ would contain just the hyperarithmetical subsets of $X$, while $\Delta$ could be some far more complicated subset of $X$.

If you require $X$ to be an element of $\mathcal A$ and require $\Delta$ to be the set of all the finitary formulas in $X$, then I believe the answer is yes, because the notion of "finitary formula" will be $\Delta_1$-definable.

I suspect you didn't mean literally what you asked, but if you did, then the answer is yes. You required only that $X$ be a subset of $\mathcal A$, so even if $X$ consists entirely of finitary sentences and $\Delta$ is all of $X$, it wouldn't follow that it's an element of $\mathcal A$.

If you require $X$ to be an element of $\mathcal A$, to avoid the situation in the preceding paragraph, the answer is still yes, since you require $\Delta$ just to be some subset of $X$ consisting of finitary formulas. So, for example, $X$ could be the set of all finitary formulas of $L$, in which case $\mathcal A$ would contain just the hyperarithmetical subsets of $X$, while $\Delta$ could be some far more complicated subset of $X$.

If you require $X$ to be an element of $\mathcal A$ and require $\Delta$ to be the set of all the finitary formulas in $X$, then I believe the answer is no, because the notion of "finitary formula" will be $\Delta_1$-definable.