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)$. EditO(2n)$.

Comment. Sorry, I edited my first answer as I misread the condition on the average degrees.

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I think the answer is no for both questions. Take a cycle on $n$ vertices, and let each edge have weight $n/2$. Then every minimum weight spanning tree has weight $n(n-1)/2$, which is $O(n^2)$.

Edit. Sorry, I misread the condition on the average degrees.