Infinite series and its upper and lower limit.

By pairing adjacent terms, you’ve actually changed the series with which you’re working: you’re working with $\sum_{n\ge 1}\left(\frac1{2^n}+\frac1{3^n}\right)=\sum_{n\ge 1}\frac{2^n+3^n}{6^n}$, the series whose $n$-th term is $\frac{2^n+3^n}{6^n}$. The ratio test works fine on this series:

$$\begin{align*} \lim_{n\to\infty}\frac{\frac{2^{n+1}+3^{n+1}}{6^{n+1}}}{\frac{2^n+3^n}{6^n}}&=\lim_{n\to\infty}\frac{2^{n+1}+3^{n+1}}{6(2^n+3^n)}\\ &=\lim_{n\to\infty}\frac{2^{n+1}}{6(2^n+3^n)}+\lim_{n\to\infty}\frac{3^{n+1}}{6(2^n+3^n)}\\ &=\lim_{n\to\infty}\frac{2^n}{3(2^n+3^n)}+\lim_{n\to\infty}\frac{3^n}{2(2^n+3^n)}\\ &=\lim_{n\to\infty}\frac1{3\left(1+\left(\frac32\right)^n\right)}+\lim_{n\to\infty}\frac1{2\left(\left(\frac23\right)^n+1\right)}\\ &=0+\frac12\\ &=\frac12\;, \end{align*}$$

which is essentially the calculation that you made. However, if you apply the ratio test to the original series, you find that if the $n$-th term is $a_n$, then

$$a_n=\begin{cases} \frac1{2^{(n+1)/2}},&\text{if }n\text{ is odd}\\\\ \frac1{3^{n/2}},&\text{if }n\text{ is even}\;, \end{cases}$$

so that

$$\begin{align*} \frac{a_{n+1}}{a_n}&=\begin{cases} \frac{1/3^{(n+1)/2}}{1/2^{(n+1)/2}},&\text{if }n\text{ is odd}\\\\ \frac{1/2^{(n+2)/2}}{1/3^{n/2}},&\text{if }n\text{ is even} \end{cases}\\\\ &=\begin{cases} \left(\frac23\right)^{(n+1)/2},&\text{if }n\text{ is odd}\\\\ \frac12\left(\frac32\right)^{n/2},&\text{if }n\text{ is even}\;. \end{cases} \end{align*}$$

Thus, $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ does not exist: the terms with odd indices are approaching $0$, but those with even indices are increasing without bound. Since the limit does not exist, the ratio test is inconclusive.

The ratio test applied to your modified series gives the correct answer because the original series is absolutely convergent; this permits you to combine the adjacent terms as you did, but since you don’t know ahead of time that the series converges, you can’t make use of the fact to show that it converges.

The even terms are a geometric series, with sum $\frac{1/3}{1-1/3}=\frac{1}{2}$. The odd terms are a geometric series, with sum 1. Because all the terms are positive, the series is absolutely convergent, so you may rearrange the odd terms to go first, then the even terms. Hence the overall series is convergent, with sum $\frac{3}{2}$.

In answer to your specific questions, the ratio test is indeed inconclusive in this case, because the limit of the ratios does not exist. That doesn't mean the series diverges, it means the test doesn't help. The supremum is infinite, the infimum is finite, so even with that refinement of the ratio test it's inconclusive.

With regards to your third question, the n-th root test is about specifically the n-th term. In this series, $\frac{1}{2^n}$ is the 2n-th term. If you group the terms pairwise as you did, then you need to take the limit of $\sqrt[n]{\frac{1}{2^n}+\frac{1}{3^n}}$.

The ratio test says that IF a certain limit is less than $1$ then a certain series converges and IF that limit is more than $1$ then that series diverges. The ratio test does not say whether the series converges or diverges in cases in which that limit does not exist (nor when it's equal to $1$). In this case, the limit in the ratio test does not exist, so the ratio test (in that form, at least) doesn't tell you whether the series converges or not. Thus the ratio test does not give you any conclusion that can contradict any other conclusion about whether the series converges.