Let $a_n$ be sequence of positive numbers.
Is that true that:
$\sum_{n=1}^{\infty} (-1)^n\cdot a_n$ converges $\implies$ $\sum_{n=1}^{\infty} a_{2n}-a_{2n+1} $ converges
$\sum_{n=1}^{\infty} (-1)^n\cdot a_n$ diverges to $\pm \infty$ $\implies$ $\sum_{n=1}^{\infty} a_{2n}-a_{2n+1} $ diverges to $\pm \infty$
On
$$S_N:=\sum_{n=1}^N(-1)^na_n=-a_1+a_2-a_3+a_4-a_5+\ldots+(-1)^Na_N$$
that the above series converges means $\;\lim\limits_{N\to\infty}S_N\;$ exists finitely, say equal to $\;S\;$, but now:
$$\Sigma_N:=\sum_{n=1}^N(a_{2n}-a_{2n+1})=(a_2-a_3)+(a_4-a_5)+\ldots+(a_{2N}-a_{2N+1})\implies$$
$$=\Sigma_N=S_{2N}+a_1\xrightarrow[N\to\infty]{}S+a_1$$
as parentheses are disposable in finite sums, and this proves both claims as you wrote them.
If in claim (2) it was given that the alternating series diverges (or doesn't converge), then it doesn't follow, for example
$$a_n:=\begin{cases}1-\frac1{n^2}\;,&n\;\text{odd}\\{}\\\;\;\;1,&n\;\text{even}\end{cases}\;\;\;\implies \sum_{n=1}^\infty (-1)^n a_n$$
diverges since $\;(-1)^na_n\rlap{\;\;\;\;/}\xrightarrow[n\to\infty]{}0\;$ , yet $\;\forall\,n\in\Bbb N,\;$ :
$$a_{2n}-a_{2n+1}=1-\left(1-\frac1{(2n+1)^2}\right)=\frac1{(2n+1)^2}\implies\sum_{n=1}^\infty(a_{2n}-a_{2n+1})\,\text{converges}$$
Both statements are true. The key is to go to partial sums. Let $S_n = \sum_{k=1}^{n}(-1)^na_n,$ $T_n=\sum_{k=1}^{n} (a_{2k}-a_{2k+1}).$ Then
$$T_n = a_1 + S_{2n+1}$$
for all $n.$ Suppose $S_n \to S \in \mathbb R.$ Then the same is true for any subsequence of $S_n.$ Hence $S_{2n+1} \to S,$ which implies $T_n \to a_1 + S.$
The proof of the second statement is the same if we use the following simple result: If $x_n \to \pm \infty,$ then the same is true for the sequence $y_n= x_n + c,$ where $c$ is a constant.