The first question has many easy answers: all you need is a space that contains a set with only one condensation point. For instance, let $S$ be any uncountable set, let $p$ be any point not in $S$, and let $X = S \cup \{p\}$. Make each point of $S$ an isolated point of $X$, and say that a set $A$ containing $p$ is a nbhd of $p$ iff $S \setminus A$ is finite. (In other words, $S$ is an uncountable discrete space, and $X$ is its one-point compactification.) Clearly $S^* = \{p\}$, and $(S^*)^* = \{p\}^* = \varnothing \ne S^*$.
I’ll modify the example a bit to answer the second question. First I’ll give $p$ more nbhds: now let $\{X \setminus A: A \subseteq S\text{ is countable}\}$ be a local base at $p$. ($X$ is now the one-point ‘Lindelöfization’ of $S$.) It’s still true that $p$ is the unique condensation point of $S$ in $X$. Since $S$ is uncountable, it can be partitioned into uncountably many countably infinite subsets, so let $S = \bigcup\{S_i:i \in I\}$, where $I$ is some uncountable index set, and each $S_i$ is countably infinite. For each $i \in I$ let $q_i$ be a distinct new point not in $X$, and let $Y = X \cup \{q_i:i \in I\}$. Basic nbhds of the point $q_i$ are the sets of the form $\{q_i\} \cup (S_i \setminus F)$ such that $F$ is a finite subset of $S_i$. (In other words, $\{q_i\} \cup S_i$ is the one-point compactification of the the countably infinite discrete space $S_i$.)
To sum up, $Y = S \cup \{p\} \cup \{q_i:i \in I\}$, and a base for $Y$ is given by $\begin{align*} \mathscr{B} = \left\{\{s\}:s \in S \right\} &\cup \left\{\{p\} \cup (S \setminus F):F \subseteq S\text{ is countable} \right\}\\ &\cup \left\{\{q_i\}\cup (S_i \setminus F):i \in I \land F \subseteq S_i\text{ is finite} \right\}. \end{align*}$ (It’s easy to check that this is a base for a topology.)
This almost works. Clearly S' = Y \setminus S: the point $p$ and the points $q_i$ ($i \in I$) are the cluster points of $S$. But each $q_i$ has a countable nbhd and therefore cannot be a condensation point of $S$, so $S^* = \{p\}$. Unfortunately, S' has no condensation points, so (S')^* is a subset of $S^*$, albeit a proper one.
The natural idea for fixing this is to give $\{q_i:i \in I\}$ a new condensation point that isn’t a condensation point of $S$. Unfortunately, any open set that contains uncountably many of the $q_i$ automatically contains uncountably many point of $S$, so this idea doesn’t work as it stands. A slightly more complicated version of it can be made to work, however.
Fix $i_0 \in I$. It’s well known that we can find in $S_{i_0}$ an uncountable family $\mathscr{A}$ of almost disjoint sets, i.e., of sets whose pairwise intersections are finite.
(For example, let $\varphi:\mathbb{Q} \to S_{i_0}$ be a bijection, for each $r \in \mathbb{R}$ let $\langle q_r(n):n \in \omega\rangle$ be a monotone increasing sequence of rationals converging to $r$ and $A_r = \{\varphi(x_r(n)):n\in\omega\}$, and take $\mathscr{A} = \{A_r:r \in \mathbb{R}\}$.)
For each $A \in \mathscr{A}$ let $q_A$ be a new point, and let $Z = Y \cup \{q_A:A \in \mathscr{A}\}$. Points of $Y \setminus \{q_{i_0}\}$ retain their basic open nbhds. For each $A \in \mathscr{A}$ the basic open nbhds of $q_A$ are the sets $\{q_A\} \cup (A \setminus F)$ for $F$ a finite subset of $A$; note that each of these sets is contained in the countable set $\{q_A\} \cup S_{i_0}$, so $q_A$ is not a condensation point of $S$. Finally, we change the topology at $q_{i_0}$ by taking its basic open nbhds to be the sets of the form $\{q_{i_0}\} \cup \bigcup\{B_A:A \in \mathscr{A} \setminus \mathscr{F}\},$ where $\mathscr{F}$ is a finite subset of $\mathscr{A}$ and $B_A$ is an open nbhd of $q_A$ for each $A \in \mathscr{A} \setminus \mathscr{F}$. Each of these nhbds also intersects $S$ in a subset of $S_{i_0}$, so $q_{i_0}$ is also not a condensation point of $S$; $q_{i_0}$ is a condensation point of $\{q_A:A \in \mathscr{A}\}$, however.
It should now be easy to see that $S^*$ is still just $\{p\}$, S' = Z \setminus S, and (S')^* = \{q_{i_0}\}, so that neither of $S^*$ and (S')^* is a subset of the other.
(I found it easiest to ignore the hint.)
Added: My original answer was intended to illustrate one possible approach to the problem, but it occurs to me that an explicit description of a space like $Z$ may be helpful. Let $\mathbb{R}^+$ be the set of positive reals and $\mathbb{R}_0^+$ the set of non-negative reals, and define $\mathbb{Q}^+$ and $\mathbb{Q}_0^+$ similarly.
Let $Z_0 = (\mathbb{Q}_0^+\times\mathbb{R}_0^+) \cup (\mathbb{R}_0^+\times\{0\})$. Let $S = \mathbb{Q}^+\times\mathbb{R}_0^+$; points of $S$ are isolated in $Z_0$.
Let $P = \mathbb{P}^+ \times \{0\}$, where $\mathbb{P}^+$ is the set of positive irrational numbers. For each $p \in \mathbb{P}^+$ let $\langle q_p(n):n \in \omega \rangle$ be a monotone increasing sequence of positive rational numbers converging to $p$ in the usual topology on $\mathbb{R}$. If $p \in \mathbb{P}^+$, the sets $B(p,n) \triangleq \{(p,0)\} \cup \{(q_p(k),0):k\ge n\}$ form a local base at $(p,0)$.
For each $r \in \mathbb{R}^+$ and each finite subset $F$ of $\mathbb{Q}_0^+$ let $B(r,F) = \{(0,r)\} \cup \left((\mathbb{Q}_0^+ \setminus F)\times \{r\}\right)$; the sets $B(r,F)$ form a local base at $(r,0)$.
For each countable $F \subseteq \mathbb{P}^+$ and each function $\varphi:\mathbb{P}^+ \to \omega$ let $B(F,\varphi) = \{(0,0)\} \cup \bigcup\limits_{p\in\mathbb{P}^+\setminus F}B(p,\varphi(p));$ the sets $B(F,\varphi)$ form a local base at $(0,0)$.
Finally, let $Z = Z_0 \cup \{p_\infty\}$, where $p_\infty$ is a new point, and for each countable subset $C$ of $S$ let $B(C) = \{p_\infty\} \cup (S \setminus C\}$; the sets $B(C)$ form a local base at $p_\infty$.
Then S' = Z \setminus S, $S^* = \{p_\infty\}$, and (S')^* = \{(0,0)\}.