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
a *--------* b * c
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?
First off, let me say that you can find the answer to this question in Sage using the nauty generator. If you're going to be a serious graph theory student, Sage could be very helpful.
count = 0
for g in graphs.nauty_geng("20 180:180"):
count = count+1
print count
The answer is 4613. But, this isn't easy to see without a computer program.
At this point, perhaps it would be good to start by thinking in terms of of the number of connected graphs with at most 10 edges. Then, all the graphs you are looking for will be unions of these. You should be able to figure out these smaller cases. If any are too hard for you, these are more likely to be in some table somewhere, so you can look them up.
Connected graphs of order n and k edges is:
n = 1, k = 0: 1
n = 2, k = 1: 1
n = 3, k = 2: 1
n = 3, k = 3: 1
n = 4, k = 3: 2
n = 4, k = 4: 2
n = 4, k = 5: 1
n = 4, k = 6: 1
n = 5, k = 4: 3
n = 5, k = 5: 5
n = 5, k = 6: 5
n = 5, k = 7: 4
n = 5, k = 8: 2
n = 5, k = 9: 1
n = 5, k = 10: 1
.
.
.
n = 10, k = 9: 106
n = 10, k = 10: 657
n = 11, k = 10: 235
I used Sage for the last 3, I admit. But, I do know that the Atlas of Graphs contains all of these except for the last one, on P7.
Best Answer
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.