Your cases are not exhaustive. You can still have a subgraph of $K_{4,4}$ that is planar and is not a subgraph of $K_{1,7}$, $K_{2,6}$, or $C_8$. Think, for example, of the graph of a cube. Also, you seem to be confusing edges and vertices. The question asks for a graph with $8$ vertices and $13$ edges, but you seem to be talking about graphs with $8$ edges.
While the statement is correct; a planar graph cannot have $8$ vertices and $13$ edges, a better approach is to use Euler's formula. Let $G$ be a planar graph with $8$ vertices and $13$ edges. Without loss of generality, $G$ is connected, so $v + f - e = 2$ (assuming $G$ is connected). This gives us $f = 7$. Now if $G$ is $2$-colorable, then every face has at least $4$ edges on it. So we can "recount" the vertices: we have $7$ faces, each with at least $4$ edges on them, and each edge appears in at most two face. So $e \ge 7 \cdot 4 / 2 = 14$, a contradiction.
Best Answer
$K_4$ is a graph on $4$ vertices and 6 edges. The line graph of $K_4$ is a 4-regular graph on 6 vertices as illustrated below:
It has a planar drawing(Hence planar):