The solution provided here is highly inelegant as it uses a lemma which is a more general result. This 'solution' should instead be considered as more of a hint: a minimal solution can instead be found by extracting the most relevant parts of the proof of the lemma.
Note that $w_1(e_1)<w_1(e_2)\Leftrightarrow w_2(e_1)>w_2(e_2)$ implies that $w_1(e_1)=w_1(e_2)\Leftrightarrow w_2(e_1)=w_2(e_2)$. In particular, if we order the edges in a non-decreasing order according to $w_1$, then we have also ordered them in non-increasing order according to $w_2$.
Now, recall that any minimal spanning tree (for $(G,w_1)$) can be found by Kruskal's algorithm (lemma -- https://cs.stackexchange.com/a/95625/109876), so in particular there is an order on the edges which is non-decreasing for $w_1$ such that $\mathcal{T}(\mathcal{V},\mathcal{E}')$ is selected. This same order is non-increasing for $w_2$, and can therefore be used to find a maximum spanning tree (for $(G,w_2)$) using Kruskal's algorithm. But since Kruskal's selection process is independent of the actual weights (and depends only on the ordering and whether the current edge creates a cycle or not), the algorithm will output $\mathcal{T}(\mathcal{V},\mathcal{E}')$ again.