Number of subgraphs with given edge parity I would like to know whether counting number of induced (full) subgraphs (of an undirected graph) that have even number of edges is P or #P-complete. Additionally, is the problem easier if we assume that given undirected graph has constant degree (all vertices have up to k incident edges)?
This problem is tractable.

Let $G = (V,E)$ be the input graph with $|V| = n$ and let $e$ and $o$ be the number of vertex-induced subgraphs of $G$ with an even and odd number of edges respectively. Of course, $e + o = 2^n$. In polynomial time, we can compute $e - o$, which I explain below.  With two equations and two unknowns, we can solve the linear system to determine the value of $e$.

From $G$, we create an instance of a counting constraint satisfaction problem (#CSP). Every vertex in $V$ is a Boolean variable and every edge in $E$ a constraint $f$ that depends on its incident vertices. The constraint $f$ evaluates to 1 unless the both vertices are assigned 1, in which case, $f$ evaluates to -1. The symmetric notation for this constraint is $[1,1,-1]$.

Now the answer to this #CSP instance is (by definition)
$$\sum_{\sigma : V \to \{0,1\}} \prod_{e \in E} f(\sigma|_{I(e)}),$$
which is a sum over all vertex subsets, the product of the outputs of every constraint $f$, where $I(e)$ is the vertices incident to $e$ and $\sigma|_{I(e)}$ is the restriction of $\sigma$ to $I(e)$.

Fix a subset of $V$. Assigned 1 to the vertices in the subset and 0 to the vertices not in the subset. If either vertex incident to the same edge is assigned 0, then this edge is not in the vertex-induced subgraph and the constraint $f$ on this edge contributes a 1 to the product, which has no effect. If both vertices incident to the same edge are assigned 1, then this edge is in the vertex-induced subgraph and the constraint $f$ on this edge contributes a -1 to the product.

Vertex-induced subgraphs with an even number of edges contribute a 1 to the sum while vertex-induced subgraphs with an odd number of edges contribute a -1. Thus, the answer to this #CSP instance is exactly $e - o$.

Since the constraint $f$ is affine, this #CSP is tractable. The polynomial algorithm for any set of affine signatures is given in The Complexity of Complex Weighted Boolean #CSP by Jin-Yi Cai, Pinyan Lu, and Mingji Xia. There may be a simpler algorithm for this particular case. If such an algorithm is known, it is probably contained in one of the references cited in the above paper.