Restricted Monotone 3CNF formula: counting satisfying assignments (both modulo $2^n$ and modulo $2$)
In any graph, the parity of the number of vertex covers is equal to the parity of the number of edge covers.

To see why, check this answer and observe how the parity of $|C|$ is equal to the parity of $\Delta_{|V|} = O_{|V|} - E_{|V|}$, which is in turn equal to the parity of $O_{|V|} + E_{|V|}$, which is the number of edge covers.

Computing the parity of the number of vertex covers is $\oplus$P-hard: therefore computing the parity of the number of edge covers is $\oplus$P-hard as well.

At least the second half of the question has been settled.