A further result has been added below:
You may want to require that distinct edges intersect in at at most one point. Without this additional assumption, even cycles can be drawn in the way you describe. (Take your A-B-C-D graph and move A and B to the edge CD, with A closer to C.)
With this extra requirement, you definitely can’t draw an even cycle.
Suppose to the contrary, and let the cycle be $v_1v_2\cdots v_{2k} v_1$. Edges $v_1v_2$ and $v_3v_4$ cross, and with the extra requirement, $v_1$, $v_2$, and $v_3$ can’t be collinear (assuming coincident vertices are also disallowed). WLOG, assume the edge $v_1v_2$ is horizontal and $v_3$ is above the line through $v_1$ and $v_2$. Then $v_4$ is necessarily below that line, $v_5$ is above, and so on. But then $v_{2k}$ and $v_3$ are on opposite sides of that line, so $v_2v_3$ and $v_{2k}v_1$ are disjoint.
Added, based on discrete’s addition to the question:
Let $G$ be a graph that can be drawn without disjoint edges. Then no vertex of $G$ can have 3 neighbors, each of which has degree at least 2.
Proof: Suppose $G$ is drawn without disjoint edges, and $v$ is adjacent to $a$, $b$, and $c$, each of degree at least $2$. We will obtain a contradiction.
[Idea of proof: The three edges $av$, $bv$, and $cv$ must all lie in one half-plane with $v$ on its boundary, and the “middle” vertex cannot have another edge that intersects the other two, as required.]
Consider the half-planes of $\mathbb{R}^2$ that are on the two sides of the line through $a$ and $v$. Except for $av$, any edge incident on $a$ or $v$ lies entirely within just one of these half-planes (the one containing its vertex that isn’t $a$ or $v$). In particular, edges $vb$ and $vc$ each lie within one of these half-planes, and so does the edge from vertex $a$ that is not the edge $av$. The latter edge must intersect both $ab$ and $ac$, so $b$ and $c$ must be in the same half-plane with respect to $av$. Similarly, $a$ and $c$ must be in the same half-plane with respect to $bv$, and $a$ and $b$ must be in the same half-plane with respect to $cv$. This is impossible. If $b$ and $c$ are in the same half-plane with respect to $av$, then either $0<\angle avb < \angle avc \le \pi$ or $0<\angle avc < \angle avb \le \pi$, and the “middle” edge of the larger angle leaves one vertex in each of the half-planes it creates.
If $G$ is a simple graph with five vertices, say $v_1,v_2,v_3,v_4,v_5$, and nine edges, then it has all edges except $v_iv_j$ for some values $i,j$. $H$ is isomorphic to $G$ because you can map $c$ to $v_i$, $e$ to $v_j$, and $a,b,d$ to the other three vertices in some order.
Best Answer
I agree with Brad on the intended interpretation of the question. This means, as he said, that there is only one graph with no edges; if you think of the entire top row of your picture as one graph, that’s it. It also means that each of the three one-edge graphs in your second row is missing a vertex. The first, for instance, should be
As Brad said, the three graphs in your bottom row are all the same graph, merely drawn differently. Thus, you should have a total of eight graphs, one with no edges, three with one edge each, three with two edges each, and one with three edges.
Now look at the hint in question (4), but apply it to these graphs: there are $3$ possible edges, $ab, bc$, and $ac$, and in any given graph on the vertex set $\{a,b,c\}$ each of these edges can be either present or absent. Thus, for each of the three edges you have a two-way choice when constructing a graph: keep the edge, or discard it. For instance, keeping $ab$ and discarding the other two yields the graph that I drew above. How many ways are there to make $3$ $2$-way choices? $2\cdot 2\cdot 2=2^3=8$, right? That’s why there are $8$ graphs on the vertex set $\{a,b,c\}$. Now try applying this logic to question (4).
Notice also that we could have counted the two-edge graphs on $\{a,b,c\}$ by asking how many ways there are to pick $2$ of the $3$ possible edges to keep. How many ways are there to pick $k$ things from a collection of $n$ different things? That idea should carry you through the rest of the questions.
Added in response to comments/questions: Yes, there are $\binom{n}2$ possible edges over $V_3$. Each pair of vertices is a possible edge, and from an $n$-element set you can make $\binom{n}2$ distinct pairs. To make a graph with vertex set $V_3$, you must choose to keep some subset of these $\binom{n}2$ possible edges. A collection of $\binom{n}2$ things has $2^{\binom{n}2}$ possible subsets, so there are $2^{\binom{n}2}$ possible graphs with vertex set $V_3$, not $2^n$. (This is the same mistake that you made in question (4).)
For question (8) you want to make graphs with $k$ edges. There are as many of these as there are ways to choose $k$ of the $\binom{n}2$ possible edges. This is not $\binom{n}k$; what is it?