How to tackle prime models and saturated models 
We define $T$ as the theory of equivalence relations
  $\{R_n\}_{n\in\mathbb{N}}$ such that $R_{n+1}$ refines $R_n$. Prove that $T$
  1) has a countable prime model; but 2) no countable $\aleph_0$-saturated
  model.

I think I understand the definitions of prime models and $\kappa$-saturated models, but I have no intuition about them. Can you explain how one should approach this exercise?
 A: Based on the discussion in the comments, I'll assume we're working with the complete theory $T$ which says that $R_0$ has infinitely many classes, and for each $n$, $R_{n+1}$ refines each class of $R_n$ into infinitely many classes. I'll give a sketch of how to analyize the countable models of $T$.
$T$ is certainly not $\omega$-categorical, simply because the formulas $R_n(x,y)$ are pairwise non-equivalent for $n\in\omega$. An $\omega$-categorical theory only has finitely many formulas up to equivalence in each tuple of variables. 
The key thing to note about a model $M\models T$ is that for any finitely many elements $a_1,\dots,a_n$, if I specify for each $a_i$ some $k_i$, then there is some $b\in M$ such that for all $1\leq i\leq n$, we have $\lnot R_{k_i}(a_i,b)$, but $R_{j}(a_i,b)$ for all $j<k_i$, unless these specifications would violate transitivity. Let's call this the extension property. This is an important step toward proving quantifier-elimination and completeness of $T$, which you would need to do in order to make my arguments below precise. 
Any theory which is not $\omega$-categorical should have a non-isolated type. In this case, there's a non-isolated $2$-type, given by by $p(x,y) = \{x\neq y\} \cup \{R_n(x,y)\mid n\in \omega\}$, asserting that $x$ and $y$ are distinct but live in the same class for all the equivalence relations. Let's denote by $R_\infty$ the equivalence relation defined by $R_\infty(x,y) \iff \bigwedge_{n\in \omega} R_n(x,y)$. 
A prime model of a theory omits all of the non-isolated types, so we see that in a prime model of $T$, the $R_\infty$-classes should all have size $1$. And in fact, given a countable model $M\models T$ such that all $R_\infty$-classes have size $1$, you can show that $M$ is prime, by embedding $M$ in any other model $N\models T$ by a "forth" argument: enumerate $M$ and embed it in $N$ one point at a time using the extension property above. It's important here that all of the $R_\infty$-classes have the minimum size ($1$), so you never have to put a new element in an $R_\infty$ class in $N$ that's already full. 
On the other hand, a countable saturated model $M$ should realize as many types as possible. In particular, for any finitely many elements $a_1,\dots,a_n$ in the same $R_\infty$ class, $M$ will realize the type saying that $x$ is distinct from all the $a_i$ but goes in the same $R_\infty$ class. This shows that all $R_\infty$ classes should be countably infinite. And indeed, given a countable model $M\models T$ in which all of the $R_\infty$ classes are countably infinite, you can argue that $M$ is $\aleph_0$-saturated, again using the extension property and quantifier elimination. 
Ok, so we know what the prime model and countable saturated model should look like, just in terms of the sizes of the $R_\infty$-classes. How do we make sure they exist? Well, you can build them by hand: Think about a complete countably branching tree, where the branching at level $n$ is the $R_n$-classes. A path through the tree denotes an $R_\infty$ class. We want to pick countably many $R_\infty$ classes to put elements in. First, for each branch at level $0$, pick a path through the tree that starts down that branch, and color it blue. Then, for each branch at level $1$, pick a path through the tree that starts down that branch, and color it blue. Repeat. After handling each level, we'll have only colored countably many branches blue. To build the prime model, put one element in each $R_\infty$-class colored blue. To build the countable saturated model, put countably many elements in each of these classes. 
Finally, note that we didn't have to pick $1$ or $\aleph_0$: we could have put any number of elements in each of the blue classes. The freedom to do this shows that $T$ has continuum-many countable models (since there are continuum-many subsets of $\mathbb{N}$, and for each subset $S\subseteq \mathbb{N}$, we can make sure that there is an $R_\infty$ class of size $n$ if and only if $n\in S$). 
