Is every countable, finite-branching LTS bisimilar to a tree? Let $L$ be a finite set of labels, and let $\mathcal{C}$ be the set of finitely-branching transition systems labeled by $L$ and with a countable set of states.  Let $\sim$ denote the bisimulation relation, and let $\mathcal{C} / \sim$ denote the set of bisimulation equivalence classes of LTS's in $\mathcal{C}$.

Q1:  Does every bisimilarity equivalence class $\alpha \in \mathcal{C} / \sim$ have a tree model?

Let $L\mu$ denote the set of closed formulas in modal $\mu$-calculus.

Q2:  True or false:  $\forall \alpha \in \mathcal{C} / \sim$, $\exists F \in L\mu$ such that models$(F) \cap \mathcal{C} = \alpha$.  Or in plain English, every bisimilarity class of countable and finitely-branching LTS's over $L$ is exactly the set of countable and finitely-branching models of some modal $\mu$-calculus formula.

An answer to either would be helpful, and answers to both would be sublime :)
Q1: Yes, every LTS is bisimilar to its unfolding, which is a tree.
Q2: No, by a cardinality argument. For instance take infinite binary trees with $L=\{a,b\}$. Each tree has countable set of states and is finitely-branching, but you have uncountably many such trees, even up to bisimilarity. However you have only countably many $\mu$-calculus formulas, so some bisimilarity classes are not captured by formulas.
Actually in this setting, a bisimilarity class can be exactly captured by a $L_\mu$ formula if and only if it is the bisimilarity class of a finite structure. The unfolding of a finite structure is called a regular tree. Any formula accepting a non-regular tree must also accept a regular tree, which is not bisimilar to it. See for instance [1] for more details.
[1] The mu-calculus and model checking.
J Bradfield, I Walukiewicz.
Handbook of Model Checking, 871-919