Is every countable, finite-branching LTS bisimilar to a tree?
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