Gerald Edgar has answered the first part.

For the second, Gerald Edgar's example shows that the range of an extended signed measure need not be closed, so we consider the question for finite signed measures. It is enough to show the result for non-atomic measures: If $\mu$ is a signed measure with atoms of arbitrarily small measure, we are done. Otherwise, $\mu$ has a finite number of atoms, and we may apply the following argument to the restriction of $\mu$ to the set where it is non-atomic.

So suppose we have a non-atomic signed measure $\mu$ and let $Z$ be a measurable set of non-zero measure. Since $Z$ is not an atom, we may choose a subset $A$ of $Z$ of non-zero measure, and with $\mu(A)$ different from $\mu(Z)$. Let $E_0$ be whichever of $A$ and $Z-A$ maximizes $|\mu(\cdot)|$. Let B be a subset of $Z-E_0$ of non-zero measure with $\mu(B)$ different from $\mu(Z)$ and let $E_1$ be whichever of B and $Z-E_0-B$ maximizes $|\mu(\cdot)|$. Define the sequence $E_0, E_1, E_2,\cdots$ inductively in this way. Let $Z_n = \cup_{i=0,\cdots,n}E_i$. Since $\mu$ is finite, $\mu(Z_n)$ converges as $n\to\infty$. In particular, $|\mu(E_n)|\to 0$.
$$
2|\mu(E_n)|\ge |\mu(E_n)| +|\mu(Z-Z_{n-1}-E_n)| \ge |\mu(Z-Z_{n-1})|,$$
so $\mu(Z_n)\to \mu(Z)$, and hence the range of $\mu$ is closed.