Both series are divergent. For the series $ 1 + \tfrac{1}{2} - \tfrac{1}{3} + \tfrac{1}{4} + \tfrac{1}{5} - \tfrac{1}{6} + \tfrac{1}{7} + \tfrac{1}{8} - \tfrac{1}{9} + \cdots $, let $ (s_n) $ be the sequence of partial sums and consider the subsequence $ (s_{3n}) $:
\begin{align*}
s_{3n} = \,\, & \left( 1 + \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{4} + \frac{1}{5} - \frac{1}{6} \right) + \cdots + \left( \frac{1}{3n - 2} + \frac{1}{3n - 1} - \frac{1}{3n} \right) \\[2mm]
\geq \,\, & \left( 1 + \frac{1}{2} - \frac{1}{2} \right) + \left( \frac{1}{4} + \frac{1}{5} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{3n - 2} + \frac{1}{3n - 1} - \frac{1}{3n - 1} \right) \\[2mm]
= \,\, & 1 + \frac{1}{4} + \cdots + \frac{1}{3n - 2} \\[2mm]
= \,\, & \sum_{k=1}^n \frac{1}{3k - 2} \\[2mm]
= \,\, & \frac{1}{3} \sum_{k=1}^n \frac{1}{k - \tfrac{2}{3}} \\[2mm]
\geq \,\, & \frac{1}{3} \sum_{k=1}^n \frac{1}{k}.
\end{align*}
This demonstrates that $ s_{3n} \geq \tfrac{1}{3} \sum_{k=1}^n \tfrac{1}{k} $ for all positive integers $ n $. Since $ \sum_{k=1}^n \tfrac{1}{k} $ is unbounded in $ n $ (this is the partial sum for the harmonic series, Example 2.4.5 of Abbott 2nd edition), it follows that
$$
(s_{3n}) \text{ is unbounded} \implies (s_n) \text{ is unbounded} \implies (s_n) \text{ is divergent}.
$$
For the series $ 1 - \tfrac{1}{2^2} + \tfrac{1}{3} - \tfrac{1}{4^2} + \tfrac{1}{5} - \tfrac{1}{6^2} + \tfrac{1}{7} - \tfrac{1}{8^2} + \cdots $, let $ (s_n) $ be the sequence of partial sums. Let's look at the subsequence $ (s_{2n}) $. For any $ m \geq 2 $, we have
$$
\frac{1}{m^2} \leq \frac{1}{m(m-1)} = \frac{1}{m-1} - \frac{1}{m} \implies -\frac{1}{m^2} \geq - \frac{1}{m-1} + \frac{1}{m}.
$$
Applying this to the subsequence:
\begin{align*}
s_{2n} = \,\, & \left( 1 - \frac{1}{2^2} \right) + \left( \frac{1}{3} - \frac{1}{4^2} \right) + \cdots + \left( \frac{1}{2n - 1} - \frac{1}{(2n)^2} \right) \\[2mm]
\geq \,\, & \left( 1 - 1 + \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{3} + \frac{1}{4} \right) + \cdots + \left( \frac{1}{2n-1} - \frac{1}{2n-1} + \frac{1}{2n} \right) \\[2mm]
= \,\, & \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2n} \\[2mm]
= \,\, & \frac{1}{2} \sum_{k=1}^n \frac{1}{k}.
\end{align*}
We may conclude that
$$
(s_{2n}) \text{ is unbounded} \implies (s_n) \text{ is unbounded} \implies (s_n) \text{ is divergent}.
$$
