Shortest walk through a given subset of edges
This is NP-hard, so it's very unlikely that a polynomial-time algorithm exists.

Given any instance $G=(V, E)$ of Hamiltonian Path, create a new graph $G'=(V', E')$ in which every vertex $v \in V$ becomes a pair of vertices $v_+, v_-$ connected by an edge in $G'$.  All of these edges should also be added to $F$.  Then for each $(u, v) \in E$, add the corresponding 4 edges $(u_-, v_-), (u_-, v_+), (u_+, v_-), (u_+, v_+)$ to $E'$ (but not to $F$).  All edges in $E'$ get weight 1.

Now if and only if there is a Hamiltonian Path in $G$, there is a path (and thus also a walk) of length $2|V|-1$ in $G'$ that passes through all the edges in $F$.  (Exercise: Prove that if a walk having length $2|V|-1$ and passing through all edges in $F$ exists in $G'$, then $G$ has a Hamiltonian Path; the other direction is easy.)  IOW, if your problem could be solved efficiently by some algorithm, then that algorithm could be used as a subroutine to solve Hamiltonian Path efficiently too, and people don't think that's very likely.