The official editorial is not asymptotically optimal and involves an advanced concept that usually does not appear in USACO Gold. I derived a relatively "low-tech" and faster solution, which I will describe below.
Solution
for the sake of convenience, let $$$\sum\limits_u^v$$$ denote the XOR of all labels on the path from node $$$u$$$ to node $$$v$$$.
Subtask 1
We want a way to efficiently calculate the XOR of every label along an arbitrary path without modifications. Note that after rooting the tree at an arbitrary root $$$r$$$, if we split a path by the LCA of its endpoints, this can be computed alongside binary-jumping LCA, but that approach can't be easily transferred to subtask 2.
The key observation is that, since XOR is an involution, if we take the XOR of $$$\sum\limits_u^r$$$ and $$$\sum\limits_v^r$$$, nodes that are on both paths will not contribute. By definition, these are exactly the common ancestors of $$$u$$$ and $$$v$$$, and almost all of them are nodes that we wouldn't want to count in the query!
Since the LCA is the only common ancestor of $$$u,v$$$ on the path between them, we can simply return $$$(\sum\limits_u^r\oplus\sum\limits_v^r)\oplus e_{\text{LCA(u,v)}}$$$ for each query. The "prefix sums" of labels can be pre-computed in $$$O(n)$$$ using a simple DFS, so the overall complexity of this subtask is $$$O(n+b(n)+q\cdot (1+c(n))$$$, where $$$b(n)$$$ is the pre-computation time of the LCA method we use, and $$$c(n)$$$ is the query
