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Qiaochu Yuan
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With regards to Q1 (and part of Q4), the numbers of the form $R(G)$ are dense in $\mathbb{R}_{\ge 0}$ even when $G$ is restricted to be abelian.

Some first results. $R(G \times H) = R(G) R(H)$ if $\gcd(|G|, |H|) = 1$. We also have $R(C_p) = \frac{2}{p}$ ($p$ is always a prime) and

$$R(C_p^4) = \frac{1}{p^4} \sum_{k=0}^4 {4 \choose k}_p = 1 + \frac{3}{p} + \frac{4}{p^2} + \frac{3}{p^3} + \frac{5}{p^4}$$

where ${n \choose k}_p$ is a Gaussian binomial coefficient.

Lemma: Let $a_1, a_2, ... $ be a sequence of positive reals such that $\lim_{n \to \infty} a_n = 0$ but such that $\sum a_n$ diverges. Then the set of sums of finite subsequences of the $a_i$ is dense in $\mathbb{R}_{\ge 0}$.

Proof. Let $r \in \mathbb{R}_{\ge 0}$ and fix $\epsilon > 0$. Choose $N$ such that $a_n < \text{min}(r + \epsilon, \epsilon)$ for all $n \ge N$. Then the partial sums starting from $a_N$ diverge but begin less than $\text{min}(r + \epsilon, \epsilon)$ and increase by at most $\epsilon$ at each step, so the conclusion follows. $\Box$

Applying the lemma to the sequence $\log R(C_p^4)$ (which satisfies the hypotheses of the lemma using the fact that $\sum \frac{1}{p}$ diverges), we conclude that the numbers of the form $R(G)$ where $G$ is a product of groups of the form $C_p^4$ for distinct primes $p$ are dense in $\mathbb{R}_{\ge 1}$, and if we allow in addition the groups of the form $C_p$, then the conclusion follows.

Q2 appears to be potentially very difficult and I have not thought about Q3.

Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741