I suppose that in the preceding part of the exercise you have shown that image of $\mu$ is $[0,\infty)$, i.e. every non-negative real number can be be obtained as $\mu(A)$ for some set $A$.
More generally, for arbitrary decreasing sequence $u_n$ we have, that every positive number can be obtained as $\sum_{n\in A}u_n$ whenever $\sum_{k=1}^\infty u_k=\infty$ and $\lim\limits_{k\to\infty} u_k=0$, see e.g. here. (The claim given in that answer is more general than this.) You have shown this for $u_n=\frac1{n+1}$ in the preceding part of this exercise, it should be easy to check that the same argument works for arbitrary such sequence.
It is obvious that $A_0=\{\emptyset\}$. Now we can show that if $x>0$ can be obtained as $\mu(A)=x$, then there is infinitely many such sets, i.e. each $A_x$ is infinite.
Indeed, let $x$ has a representation $\sum_{n\in A} u_n$ and $n_0\in A$. Then we can simply omit $u_{n_0}$ from the sequence and apply the same observation to the new sequence. Again, $x$ has some representation, but it has to be a different one, since it does not contain $n_0$.
We can continue like this - we choose some $n_1$, then we obtain a new representation which does not contains $n_0$, $n_1$ etc. We will get infinitely many sets belonging to $A_x$.
However, I don't see what cardinality will the set $A_x$ have. (Aside from obvious estimate that it is between $\aleph_0$ and $\mathfrak c=2^{\aleph_0}$.)