If $\kappa$ is an inaccessible cardinal then the tree property at $\kappa$ is equivalent to weak compactness of $\kappa$, which implies that $\square(\kappa)$ fails---that is, that every coherent sequence of clubs of length $\kappa$ can be threaded.

I am wondering about other implications involving square and the tree property, namely:

If $\kappa$ is an inaccessible cardinal and $\square(\kappa)$ fails, must $\kappa$ have the tree property (and therefore be weakly compact?)

If $\kappa$ is a regular cardinal, does $\neg \square(\kappa)$ imply that $\kappa$ has the tree property?

If $\kappa$ is a regular cardinal with the tree property, must $\square(\kappa)$ fail?

(By the way, I am aware of the relative consistency result that if $\square(\kappa)$ fails for some regular cardinal $\kappa$, then $\kappa$ is weakly compact in $L$ and so in particular it has the tree property in $L$.)