None of them are necessarily true.
We can easily compute a series from its partial sums, so let's specify the $s_k$.
Define
$$
s_k=\left\{\begin{array}{}
-\frac1k&\text{if $k$ is odd}\\[4pt]
-\frac1{k^2}&\text{if $k$ is even}
\end{array}\right.
$$
Then $a_1=-1$ and for $k\gt1$,
$$
a_k=\left\{\begin{array}{}
\frac1{(k-1)^2}-\frac1k&\text{if $k$ is odd}\\[4pt]
\frac1{k-1}-\frac1{k^2}&\text{if $k$ is even}
\end{array}\right.
$$
Show that this series is not absolutely convergent, its sum is $0$, and it fails to satisfy any of the conditions.
Let $b_n = (|a_n| + a_n)/2$ and $c_n = (|a_n|-a_n)/2.$
Then the partial sums satisfy
$$\sum_{n=1}^m a_n = \sum_{n=1}^m b_n - \sum_{n=1}^m c_n, \\ \sum_{n=1}^m |a_n| = \sum_{n=1}^m b_n + \sum_{n=1}^m c_n.$$
If $\sum a_n$ converges and $\sum |a_n|$ diverges, then both $\sum b_n$ and $\sum c_n$ diverge, since
$$2\sum_{n=1}^m b_n = \sum_{n=1}^m |a_n| + \sum_{n=1}^m a_n, \\ 2\sum_{n=1}^m c_n = \sum_{n=1}^m |a_n| - \sum_{n=1}^m a_n,$$
and the sum or difference of a divergent and convergent series is divergent.
Furthermore, we have divergence to $+\infty$ in each case, as the partial sums of $|a_n|$ form a non-negative, non-decreasing sequence.
Note that
$$\{b_n: n \in \mathbb{N}, b_n \neq 0\} = \{a_n: n \in A^+, a_n \neq 0\}, \\ \{c_n: n \in \mathbb{N}, c_n \neq 0\} = \{-a_n: n \in A^-\}, $$
and it easily shown that
$$ \sum_{n\in A_+} a_n=\sum_{n=1}^\infty b_n = +\infty\\ \sum_{n\in A_-} a_n = -\sum_{n=1}^\infty c_n = - \infty
$$
Best Answer
It helps to reformulate the assumptions in terms of $S_k$. We are told that
and nothing else. Of course, there is nothing here that implies $S_k$ being zero, or positive for infinitely many values of $k$. The examples $S_k=(2+(-1)^k)/k$ and $S_k=(-2+(-1)^k)/k$ take care of all four parts, confirming Did's answer in a comment: "(a) and (b) are false in general while (c) and (d) hold".