Abundancy index and non-solvable finite groups

I can answer Questions 1 and 4.

Make sure you look at S. Carnahan's answer. It deals with Questions 2 and 3.

Questions 1 and 4: If a finite group $G$ is not solvable then $|G|$ is divisible by $|G_0|$ for some finite simple group $|G_0|$. By the CFSG, either $12\bigm||G_0|$ or $G_0$ is a Suzuki group. If $12\bigm||G_0|$ then $$\frac{\sigma(|G|)}{|G|}\geq\frac{\sigma(|G_0|)}{|G_0|}>\frac{\sigma(12)}{12}=\frac{7}{3}.$$ If $G_0$ is a Suzuki group then $320\bigm||G_0|$ and $$\frac{\sigma(|G|)}{|G|}\geq\frac{\sigma(|G_0|)}{|G_0|}>\frac{\sigma(320)}{320}>\frac{7}{3}.$$ Thus, every non-solvable group has abundancy index larger than $\frac{7}{3}$.

As I mentioned in a comment, Question 2 (in its revised form) has a negative answer, because odd natural numbers have unbounded abundancy index, while the Odd Order Theorem implies all groups of odd order are solvable.

Weaker version 2 has a positive answer: If $\beta$ is sufficiently close to 1, then any $n > 1$ whose abundancy index is greater than $\beta e^\gamma \log \log n$ is a multiple of 60, so there is a group of order $n$ that is unsolvable.

As I mentioned in a different comment, Question 3 is true subject to well-known open conjectures, such as [Dickson's conjecture][1]. In particular, it suffices to show that there are infinitely many primes $p$ such that $p-1$ is 4 times a prime and $p+1$ is 6 times a prime. [1]: https://en.wikipedia.org/wiki/Dickson%27s_conjecture