$\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}\;\;(\text{mod}\;p)$

let $f_{a,p} : \Bbb Z/p\Bbb Z \to \Bbb Z/p\Bbb Z $ given by $f(x) = x^2-a$.
Then $x \in S(a,p)$ if and only if there is some $n$ such that $f_{a,p}^{\circ n}(x) = x$,.

Draw a graph where the vertices are the elements of $\Bbb Z/p\Bbb Z$ and where there is an arrow from $x$ to $f_{a,p}(x)$. Then $S(a,p)$ is the set of vertives that are parts of cycles.

Iterating $f_{a,p}$ on any element eventually ends up in a cycle, so $S_{a,p}$ is non empty.

If $p=2$ then $S(a,p)$ is a bijection so every element is part of a cycle.

If $p \neq 2$, then every element $x \neq -a$ has either $0$ or $2$ preimages $y_1,y_2$, and only one of them can be part of a cycle (since $x$ can only appear once in the cycle it will have either $\ldots \to y_1 \to x \to \ldots$ or $\ldots \to y_2 \to x \to \ldots$), so $S(a,p)$ has at most $(p+1)/2$ elements (and so the cycle lengths can be at most $(p+1)/2$)

This, for now in more like an extended comment (answering Conjecture 2, but not yet quite Conjecture 1.)

I just copy here some of my observations from the comments. I believe both conjectures are true, except perhaps the bound $p-1$ when $p=2$ (but I don't care much of this exception and didn't think in detail about it).

If $x\in S(a,p)$ where $p$ is an odd prime, then there is a chain of at most $(p+1)/2$ radicals. This is because there are at most $(p+1)/2$ quadratic residues $\mod p$ so if the chain were longer there ought to be some repetition in it, so the chain could be shortened. This shows that Conjecture 2 is correct for odd primes $p$ since in this case $(p+1)/2\le p-1$.

Re conjecture 1, it is true too. Start with any $x_0$ and iterate $x_{n+1}=\sqrt{a+x_n}$ ... I see a gap in my argument though. Assuming the square root "exists often enough", then there will be some repetition in this chain, i.e. $x_m=x_{m+k}$ for some positive integer $k$, showing that $x_m\in S(a,p)$. But I need to think of the square root "exists often enough" part of my "answer"

Edit. Aah, mercio posted an answer, there is a cycle. One need not worry about the existence of the square root, since one could find a cycle for the square function (rather than for the square root function), and once we have a cycle for the square function, then it is also a cycle for the square root function, obviously, if we walk the opposite direction. (More precisely, it is not exactly the square function, but $f(x)=x^2-a$ as in mercio's answer, as also observed just slightly later by Jyrki Lahtonen in the comments).