We have the following for $n \in \mathbb{N}$ :
$$r(a^n)-r(a^n-1)=
\begin{cases}
2a^n-1-\sigma(a^n) & \text{if $2 \mid a$} \\
\frac{3a^n-1}{2}-\sigma(a^n) & \text{if $2 \nmid a$}
\end{cases}$$
Question $1$
We have:
$$r(2^n)-r(2^n-1)=2(2^n)-1-\sigma(2^n)=2^{n+1}-1-(2^{n+1}-1)=0$$
$$r(3^n)-r(3^n-1)=\frac{3(3^n)-1}{2}-\sigma(3^n)=\frac{3^{n+1}-1}{2}-\bigg(\frac{3^{n+1}-1}{2}\bigg)=0$$
This solves Question $1$.
Question $2$
As mathlove points out, this is false. $110$ is a counterexample. A weaker version of this can be proven:
All values of $a$ which are not powers of $2$ or powers of $3$ lead to $r(a^n)-r(a^n-1)$ eventually being all positive or all negative.
I personally feel that this is the heart of the problem. We have:
$$a=\prod_{i=1}^k p_i^{a_i} \implies \sigma(a^n)=\prod_{i=1}^k \frac{p_i^{a_in+1}-1}{p_i-1}$$
This gives:
$$\frac{\sigma(a^n)}{a^n}=\prod_{i=1}^k \frac{p_i-(1/p_i^{a_in})}{p_i-1} \implies \lim_{n \to \infty} \frac{\sigma(a^n)}{a^n}=\prod_{i=1}^k \frac{p_i}{p_i-1}=c_a$$
Now, as $n$ gets large, we can see that $\frac{\sigma(a^n)}{a^n} \to c_a$ which is a constant for $a$. We know:
$$r(a^n)-r(a^n-1)=
\begin{cases}
2a^n-1-\sigma(a^n) & \text{if $2 \mid a$} \\
\frac{3a^n-1}{2}-\sigma(a^n) & \text{if $2 \nmid a$}
\end{cases}
$$
This gives:
$$r(a^n)-r(a^n-1) \sim
\begin{cases}
2a^n-c_a a^n & \text{if $2 \mid a$} \\
\frac{3}{2}a^n-c_aa^n & \text{if $2 \nmid a$}
\end{cases}
$$
It is clear that this value will eventually become all positive or all negative (which is what we are required to show) when $c_a \neq 2$ and $c_a \neq \frac{3}{2}$.
If $c_a=2$, we have:
$$\prod_{i=1}^k \frac{p_i}{p_i-1}=2$$
Clearly, we need an even prime, so let $p_1=1$. Then:
$$\prod_{i=2}^k \frac{p_i}{p_i-1}=1$$
But this forces there to be no other primes, since $\frac{p_i}{p_i-1}>1$ for all primes. Since $2$ is the only prime divisor of $a$, $a$ must be a power of $2$ (which is one of the exceptions).
The same argument works for $\frac{3}{2}$ where $p_1=3$ is forced to be the only prime, giving powers of $3$ as an exception. Thus, these are the only exceptions (and we also know that in these cases, the difference is always zero). Hence, we have proven the required.
Thus, we have proven the following lemma-
Define: $$c_a=\prod_{p \mid a} \frac{p}{p-1}$$
The difference $r(a^n)-r(a^n-1)$ is-
- eventually all negative, if $c_a>2$.
- all $0$, if $c_a=2$ (which only happens for $a=2^t$).
- eventually all negative, if $c_a>1.5$ and $a$ is odd.
- all $0$, if $c_a=1.5$ (which only happens for $a=3^t$).
- eventually all positive, if $1.5>c_a$.0
Question $3$
Let $a=2p \space (p>3)$. Then:
$$c_a=\frac{2}{1} \cdot \frac{p}{p-1} > 2$$
Thus, $r(a^n)-r(a^n-1)$ is eventually all negative (from lemma). At $n=1$,
$$r(a)-r(a-1)=4p-1-\sigma(2p)=p-4>0$$
Thus, there is an eventual transition from positive to negative.
Let $a=3q \space (q>7)$. Then:
$$c_a=\frac{3}{2} \cdot \frac{q}{q-1} > \frac{3}{2}$$
Also, $a$ is odd. Thus, $r(a^n)-r(a^n-1)$ is eventually all negative (from lemma). At $n=1$,
$$r(a)-r(a-1)=\frac{9q-1}{2} - \sigma(3q) = \frac{q-9}{2}>0$$
Thus, there is an eventual transition from positive to negative.
Question $4$
We are to simply substitute $a=p$, $a=p^2$ and $a=pq$.
If $a=p$,
$$r(a^n)-r(a^n-1)=\frac{3p^n-1}{2}-\sigma(p^n)=\frac{3p^n-1}{2}-\frac{p^{n+1}-1}{p-1}>0$$
If $a=p^2$,
$$r(a^n)-r(a^n-1)=\frac{3p^{2n}-1}{2}-\sigma(p^{2n})=\frac{3p^{2n}-1}{2}-\frac{p^{2n+1}-1}{p-1}>0$$
If $a=pq$,
$$r(a^n)-r(a^n-1)=\frac{3p^nq^n-1}{2}-\sigma(p^nq^n)=\frac{3p^nq^n-1}{2}-\bigg( \frac{p^{n+1}-1}{p-1} \cdot \frac{q^{n+1}-1}{q-1} \bigg)>0$$
All of these can be shown by expanding (strenuous but straightforward). This completes all $4$ problems.
Best Answer
$146$ can be written as squares of two primes: $$146=5^2+11^2=5^2+(1+4+6)^2=(1+4)^2+(1+4+6)^2.$$