There are exactly four ways we could attach $1$ to a connected graph on $4$ vertices using one edge.
(that sentence should read less obviously than it is. if we only have one edge, which do we want it to go to? there are four options.)
There are 12 ways we could attach $1$ to a connected graph on $4$ vertices using two edges.
First we select one vertex, then we select a different vertex. Thus 4*3=12 ways
Similarly, there are $4*3*2$ ways to connect $1$ to the connected graph on $4$ vertices using $3$ edges, and there are also $4*3*2$ ways to do it using four edges.
(In general, the formula would be ${4\choose k}k!=\frac{4!}{(4-k)!}$, because it's the number of ways to pick which vertices you connect $1$ to, multiplied by the number of ways they can be permuted.)
Wait, crap I'm double counting. We don't need to permute them. There's just ${4 \choose k}$ ways for each connection, so, $4, 6, 4, 1$ ways.
The reason that's important is that, because the connection to $1$ has no bearing on whether or not the graph is connected, we can just let $C(4)$ be the number of connected graphs on four vertices, and we know there are at least $(4+6+4+1)C(4)$ connected graphs on five vertices.
So, that takes care of the connected case.
For the disconnected case, let's start with $k=4$.
Any disconnected graph will become connected if it is connected up to $1$ by $k=4$ edges.
Now, for $k=3$, the three points that $1$ connects to must be disconnected without $1$. There are only $4$ disconnected graphs on three vertices. There is one vertex $v$ which is not connected to $1$, so it must connect to the three vertices adjacent to $1$ also.
There are six ways for it to connect to the empty graph on three vertices, then for each of the three disconnected graphs with one edge there is one option, so that is nine ways.
So for $k=3$ we'll have 9*4=36 disconnected graphs.
For $k=1$ we're just counting connected graphs on three vertices (there are four) because we need to connect an isolated point to a connected set of three vertices.