is there a certain number of cut sets of a graph with n vertices? Firstly, how many cut sets do graphs have? Do they have more than one ? If so, is there a way of knowing how many of them there are? Also, what does it mean for a disconnecting set of edges to not be a cut set? Isn't disconnecting the graph the main definition of a cut set ? 
 A: A cut set is the set of edges between the two parts of a bipartition of the nodes in a graph.  So if a graph has $n$ nodes, there are $2^{n-1} - 1$ ways to partition those nodes into two nonempty subsets (bipartitions).
Unfortunately the number of cut sets is not quite in one-to-one correspondence with those bipartitions (which we easily counted).  In particular, if the graph $G$ on $n$ nodes is disconnected, there can be various bipartitions that give us the same empty cut set.
So $2^{n-1} - 1$ is an upper bound on the number of cut sets, but it is not tight.  Consider the examples of graphs on four nodes, and see if you can find a pattern for connected graphs versus disconnected graphs.
While every cut set is a "disconnecting set of edges" (removing them guarantees a disconnect graph will result), not every disconnecting set of edges can be realized as a cut set.  This is true even if we assume graph $G$ is connected.  Again you should be able to pick out an example of this using graphs on four nodes.
A: Edit: this hint may make most sense if you have studied the spectral clustering connection to graph cuts.

To build some intuition on this consider the masking matrix $${\bf M}= {\bf I}_n\otimes {\underset{m\times m \text{ block of 1:s}}{\underbrace{({\bf 1}_m{{\bf 1}_m}^T)}}}$$
Where $\otimes$ is Kronecker product. Introduce a random matrix ${\bf R}=uniform([0,1],m\times n)$. Then we reset ${\bf R = I+R+R}^T$ to ensure a symmetric matrix with large diagonal elements which is what we would expect from a candidate affinity matrix.
We can now consider $\bf A = M \circ R$ ($\circ$ Schur/Hadamard product) being a fake affinity matrix of a block-diagonal matrix from which we can construct a normalized Laplacian (${\bf D} = \text{diag}(\text{diag}({\bf A}))$):
$$\bf L = D^{-1/2}AD^{-1/2}-I$$
Now maybe you have some ideas of how to proceed?

EDIT Ok maybe my hint was a bit cryptic. The next step is to calculate and sort eigenvalues and check corresponding eigenvectors. As happens the $n$ largest eigenvalues (which will be all eigenvalues $>0$) will correspond to eigenvectors which are non-zero for the $m$ elements uniquely corresponding to a particular $m\times m$ block and zero everywhere else. If you are still unsure what it means, you can experiment with switching the order of the Kronecker factors where $\bf M$ is constructed and see what happens to those eigenvectors.
