And lastly: Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?
I think that we cannot.
In the following, I'll explain the reason why I think that we cannot.
In the answer, I wrote
"If you get an improved lower bound for $f(k)$, then you can say that there is an integer $a$ such that $k\le a$ "
where $f(k)=I(q^k) + I(n^2)$.
Also, in a comment below the answer, I wrote
"if you get an improved lower bound $g(q)$, then from the above fact, we see that there is only one $k=k_0$ such that $f(k)=g(q)$. Then, we have $k\le \lfloor k_0\rfloor$ "
where "the above fact" is the fact that $f(k)$ is decreasing.
In short, we can say that there is an integer $a$ such that $k\le a$ if we get a function $g(q)$ (which is a fuction only on $q$) such that
$$I(q^k)+I(n^2)\ge g(q)\gt 3 - \frac{q-2}{q(q - 1)}$$
Now, let us see what you've got.
Let $$G(k,q)=\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$
You have
$$I(q^k) + I(n^2) > G(k,q)\gt 3 - \frac{q-2}{q(q - 1)}$$
which is correct since
$$I(q^k) + I(n^2)-G(k,q)=\frac{ (q- 4)q^{k+2}+2 q^{k + 1} + (3 q - 4 )q + 2}{q^{k+2}(q - 1) (q^{k + 1} - 1) (q^{k + 1} + 1)}\gt 0$$
$$G(k,q)-\bigg(3 - \frac{q-2}{q(q - 1)}\bigg)=\frac{ (q - 4) q^{k + 2} +2 q^{k + 1} + 2( q - 2 )q + 2}{q^{k+2}(q - 1) (q^{k + 1} + 1)}\gt 0$$
Also, we have
$$\frac{\partial G(k,q)}{\partial k}=-\frac{ (q- 4) q^{2 k + 3}+ 2 q^{2 k + 2}+ 4( q- 2)q^{k + 2} +4 q^{k + 1} + 2( q - 2) q + 2}{q^{k+2}(q - 1) (q^{k + 1} + 1)^2}\ln q\lt 0$$
$$\lim_{k\to\infty}\bigg(I(q^k) + I(n^2) \bigg)=\lim_{k\to\infty}G(k,q)= 3 - \frac{q-2}{q(q - 1)}$$
These imply that you don't get a function $g(q)$ (which is a fuction only on $q$) such that
$$I(q^k)+I(n^2)\ge g(q)\gt 3 - \frac{q-2}{q(q - 1)}$$
So, I think that you cannot say that there is an integer $a$ such that $k\le a$.
Added :
I'm going to add another explanation with graphs.
Let $$f(k)=I(q^k) + I(n^2),\qquad G(k)=\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}},$$
$$Q=3 - \dfrac{q-2}{q(q - 1)}$$
where $q,Q$ are seen as constants.
Since we see that $f(k)\gt G(k)\gt Q$ holds for all $k,q$ and that both $y=f(k)$ and $y=G(k)$ are decreasing with $\displaystyle\lim_{k\to\infty}f(k)=\displaystyle\lim_{k\to\infty}G(k)=Q$, we see that their graphs are as follows :
Now, if you take a fixed positive integer $k=k'$, then we have the following :
This explains why you cannot say that
$$f(k)\gt G(k′)$$
holds for all $k$.
Added 2 :
I'm going to add more explanations with steps :
(1) $f(k)\gt G(k)\gt Q$ holds for all $k,q$.
(2) Both $y=f(k)$ and $y=G(k)$ are strictly decreasing.
(3) $\displaystyle\lim_{k\to\infty}f(k)=\displaystyle\lim_{k\to\infty}G(k)=Q$.
(4) If we let $k=k'$ where $k'$ is a fixed positive number larger than $1$, then $G(k')$ is entirely in terms of $q$ alone.
(5) $Q\lt G(k')\lt f(1)$
(6) It follows from $(2)(3)$ that, for any $Y$ satisfying $Q\lt Y\lt f(1)$, there exists only one $k$ such that $f(k)=Y$.
(7) It follows from $(5)(6)$ that there exists only one $k$ such that $f(k)=G(k')$. Let $k_0$ be this $k$.
(8) For $k\lt k_0$, we have $f(k)\gt G(k')$.
(9) For $k\ge k_0$, we have $f(k)\le G(k')$.
(10) It follows from $(8)(9)$ that $f(k)\gt G(k')$ does not hold for all $k$.
Added 3 :
You seem to think that the following claim is true :
Claim : If there are two functions $f(x),g(x)$ such that
$$f(x)\gt g(x)\gt 1$$
holds for all $x\gt 0$, then there is a fixed positive integer $m$ such that $f(x)\gt g(m)$ holds for all $x\gt 0$.
This claim is false. Take $f(x)=\dfrac 1x+1$ and $g(x)=\dfrac{1}{x+1}+1$ for which $f(x)\gt g(x)\gt 1$ holds for all $x\gt 0$, but $f(x)\gt g(m)$ does not hold for $x\ge m+1$.
Best Answer
This is an answer to the question (1).
Let $$f(k):=\varphi(q^k)-D(q^k)=q^{k-1}(q-1)-\bigg(2q^k-\frac{q^{k+1}-1}{q-1}\bigg)=\frac{q^{k-1}-1}{q-1}$$ Then, we see that $f(k)$ is increasing.
So, if $k\gt 1$ and $k\equiv 1\pmod 4$, then we have $f(k)\ge f(5)=\dfrac{q^{4}-1}{q-1}$ which is equivalent to $$D(q^k)\le\varphi(q^k)-\frac{q^4-1}{q-1}$$ Using $$D(n^2) < \frac{2n^2}{q}$$ we get $$D(q^k)D(n^2) < \frac{2n^2}{q}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$ Dividing the both sides by $2q^k n^2$, we obtain $$\frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$ which can be written as $$3 - \bigg(I(q^k) + I(n^2)\bigg) < \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg)$$
We therefore finally have $$3 - \frac{1}{q^{k+1}}\bigg(\varphi(q^k)-\frac{q^4-1}{q-1}\bigg) < I(q^k) + I(n^2) $$