I started this, went to dinner, and came back to see that David Speyer anticipated me, but I'll put it in anyway. Here is a theorem, a story and a result somewhat along the lines you want. 

Theorem: If $\mathcal{F}$ is a forest with $n$ (labelled) vertices and $k$ connected components of sizes $n_1,n_2,\cdots,n_k$ then the number $T(\mathcal{F})$ of completions to a labelled tree is $n_1n_2\cdots n_kn^{k-2}.$

The proof is left to the interested reader but it is an easy induction on $k \ge 1$ and includes at the other extreme of $k=n$ the usual enumeration of labelled trees.

Story: I discovered this as an undergraduate. Of course I knew that there are like 100 proofs of the $n^{n-2}$ theorem but I thought the induction on the number of undetermined edges  *might* be a slightly new twist. I had the occasion to show it to Frank Harary who said: "This is well written and deserves a place in the literature, I am editing a new journal, submit it." So I did. A few months later it came back with a letter from Haray: "I got the referees report, never submit this anywhere again!" and I didn't (until now!).

So for your Q2: Label an edge from $v_i$ to $v_j$ as $e_{i,j}$ (with $i \lt j$). Then in the Laplacian matrix if you plug in  $n+\sum_j e_{i,j}$  instead of $\deg(v_i)$ and $-1-e_{i,j}$ instead of -1 when that edge connects vertices $i$ and $j$, you get a modified combinatorial Laplacian. Taking the determinant of any minor of this matrix gives a modified Kirchhoff polynomial which is a weighted enumeration of the spanning forests of the graph, where each term is a monomial contains the variables for all the edges in a given forest $\mathcal{F}$ and the coefficient is $T(\mathcal{F}).$ So this polynomial spits out all the spanning forests including the empty one (times $n^{n-2}$) ,each of the spanning trees, and everything in between.