I think the answer is no for both questions. Take a cycle Let $T$ be the unique tree on $n$ 2n$ vertices , with two adjacent vertices $u$ and let each edge have $v$ of degree $n$. Let $e=uv$. Let the weight of $n/2$. Then every minimum e$ be $n^2$ and all other edges to have weight spanning tree 0. Then the sum of all the average weights is
$n^2/n + n^2/n = 2n = |V(T)|$.
However, $T$ has total weight $n(n-1)/2$, n^2$, which is not $O(n^2)$.
Comment. Sorry, I edited my first answer as I misread the condition on the average degrees.