For the first series I applied the criterion of the report and discussed after A. Arriving at the fact that the limit of the report is | a |. When I replaced x with $\frac{-1}{a}$, respectively $\frac{1}{a}$. I'm not sure I found what I needed.
For the second series I applied the root criterion and obtained the radius = $ \frac{1}{3} $. But in the end I don't think it went well! Thank you very much !
If $ x $ is a real such that $ 0<x<1 $, then $ \prod\limits_{n\geq 1}{\left(1+x^{n}\right)} $ converges, simply because the sequence $ \left\lbrace\prod\limits_{k=1}^{n}{\left(1+x^{k}\right)}\right\rbrace_{n} $ is increasing, and bounded : $ \left(\forall n\in\mathbb{N}\right),\ \prod\limits_{k=1}^{n}{\left(1+a^{k}\right)}\leq\prod\limits_{k=1}^{n}{\exp{a^{k}}}=\exp{\left(\frac{a-a^{n+1}}{1-a}\right)}\leq\exp{\frac{a}{1-a}}\cdot $
Thus, for any $ 0<x<1 $, we'll be denoting $ C_{x}=\prod\limits_{n=1}^{+\infty}{\left(1+x^{n}\right)} \cdot $
Let's find out the radius of convergence of the first series when $ a<1 : $
We have : $$ \frac{a^{n}}{\prod\limits_{k=1}^{n}{\left(1+a^{k}\right)}}\underset{n\to +\infty}{\sim}\frac{a^{n}}{\prod\limits_{n=1}^{+\infty}{\left(1+a^{n}\right)}}=\frac{a^{n}}{C_{a}} $$
That means the radius of the power series $ \sum\limits_{n\geq 1}{\frac{a^{n}}{\prod\limits_{k=1}^{n}{\left(1+a^{k}\right)}}x^{k}} $ would be exactly the radius of $ \sum\limits_{n\geq 1}{a^{n}x^{n}} $ which is $ \frac{1}{a} \cdot $
If $ a>1 $, then $ \prod\limits_{k=1}^{n}{\frac{1+a^{k}}{a^{k}}}\underset{n\to +\infty}{\longrightarrow}C_{\frac{1}{a}} $, thus : $$ \prod_{k=1}^{n}{\left(1+a^{k}\right)}\underset{n\to +\infty}{\sim}C_{\frac{1}{a}}\prod_{k=1}^{n}{a^{k}}=C_{\frac{1}{a}}a^{\frac{n\left(n+1\right)}{2}} $$
Hence, $$ \frac{a^{n}}{\prod\limits_{k=1}^{n}{\left(1+a^{k}\right)}}\underset{n\to +\infty}{\sim}\frac{a^{n}}{C^{\frac{1}{a}}\prod\limits_{k=1}^{n}{a^{k}}}=\frac{1}{C_{\frac{1}{a}}}a^{\frac{n\left(1-n\right)}{2}} $$
That means the radius of the power series $ \sum\limits_{n\geq 1}{\frac{a^{n}}{\prod\limits_{k=1}^{n}{\left(1+a^{k}\right)}}x^{k}} $, in this case, would be exactly the radius of $ \sum\limits_{n\geq 1}{a^{\frac{n\left(1-n\right)}{2}}x^{n}} $ which is the limit of the ratio $ \frac{a^{\frac{n\left(1-n\right)}{2}}}{a^{\frac{\left(n+1\right)\left(1-n-1\right)}{2}}}=a^{n}\underset{n\to +\infty}{\longrightarrow}+\infty \cdot $
If $ a=1 $, the series would be geometric, and would have a radius of convergence being equal to $ 2 \cdot $
What you have done for the second series seems correct to me.