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(Caveat: normally I wouldn't answer a question with such a limited knowledge of the general theory, but classical analytic number theory seems not so well represented by active MO members.)
Suppose that $A$ is a finite abelian group. Then I claim that given any set of at least $|A| + 1$ (not necessarily distinct) elements of $A$ one can find a proper subset whose sum is the identity. Proof: Denote the elements $a_i$ for $i = 1$ to $|A|+1$. By the pigeonhole principle, either one of the $|A|$ sums $\sum_{i=1}^{r} a_i$ for $r = 1$ to $|A|$ is the identity, or two of the sums are the same element of $|A|$, in which case consider the difference. We deduce from this the following: Let $n$ be any integer coprime to $a$ with more than $r:=|(\Z/a \Z)^{\times}|$ prime factors. Then $n$ has a proper divisor of the form $1 \mod a$.
Suppose that $k$ cannot be represented by the form $amn \pm m \pm n$, and suppose that $a > 2$. It is simple to deduce that this is equivalent to asking that $ak+1$ and $ak-1$ have no proper divisors of the form $\pm 1 \mod a$. It follows that $ak+1$ and $ak-1$ each have at most $r=|(\Z/a \Z)^{\times}|$ prime
factors. The integers with at most $r$ prime factors
are sometimes called $r$-almost primes. If $\pi_r(x)$ counts the number of $r$-almost
primes $\le x$ then
$$\pi_r(x) \sim \frac{x (\log \log x)^{r-1}}{\log(x)}.$$
(Compare this to the prime number theorem when $r = 1$.) In particular, we see that the
$r$-almost primes have zero density (in any sense), and thus:
2) The density of integers that can not be represented in the form $amn + m + n$ is zero.
Similarly, the density of integers that cannot be represented in the form $amn + m -n$ is zero. In particular, the density of the $a$-asterios numbers, the integers that can neither
be represented in the form $amn+m+n$ nor $amn+m-n$, is zero.
Let $\pi_{r,2}(x)$ denote the number of twin $r$-almost primes less than or equal to $x$, that is, the number of integers $n \le x$ such that $n$ and $n+2$
are both $r$-almost primes.
(For example, $\pi_{1,2}(x)$ counts the number of twin primes less than $x$.)
What do we know about this function? Brun was the first to give an upper bound
for $\pi_{r,2}(x)$ using sieving techniques. Refinements by others (in particular
Selberg) allowed one to obtain the estimate
$$\pi_{1,2}(x) \ll \frac{x}{(\log x)^2},$$
which gives the correct (conjectural) order of magnitude.
Without going into the Selberg sieve, let me say that what these arguments
really
give is decent upper and lower bounds of the following kind (for large $x$):
$$\frac{A x}{(\log x)^2} <
\left\{n < x, \ p \nmid n(n+2) \ \text{if} \ p < x^{\alpha}\right\}
< \frac{B x}{(\log x)^2}$$
for non-zero constants $A$ and $B$,
where $0 < \alpha < 1$ is some fixed small constant, which we might imagine for the
sake of argument is $1/10$. Since every twin prime $>x^{\alpha}$ contributes to
this sum, this gives the correct (up to a constant) upper bound for
$\pi_{1,2}(x)$. It also gives a lower bound for $\pi_{10,2}(x)$, since if every
factor of $n < x$ is at least $x^{1/10}$, then $n$ has at most $10$ prime factors.
The motto I learnt about sieving was the following: upper bounds are easy, lower bounds are hard. Thus, since we are interested in bounding $\pi_{r,2}(x)$, it seems that we are in good shape.
However, there is a subtlety here. Let $\pi(x,z)$ denote the number of integers $n$ less
than $x$ such that every prime factor of $n$ is at least $z$. It's clear by the
argument of the last paragraph that
$\pi_r(x) \ge \pi(x,x^{1/r})$. One might imagine that these numbers are roughly of the
same magnitute. However, it turns out that $\pi_r(x)$ is much bigger than $\pi(x,x^{1/r})$.
The latter is comparable to the number of primes less than $x$, where the former has
an extra factor of $(\log \log x)^{r-1}$. The reason is that $\pi_r(x)$ is dominated
by numbers with (a few) small prime factors. In fact,
as Kowalski pointed out to me, it is not even obvious that
one can easily obtain the
correct upper bound for $\pi_r(x)$ simply by sieving over primes.
From the asymptotic for $\pi_{r}(x)$, one expects that
$$(*): \qquad \pi_{r,2}(x) =^{?} \ O\left(\frac{x (\log \log x)^{2r-2}}{(\log x)^2}\right).$$
(EDIT: My resident expert reports that this is known. Here is a sketch of the idea in the simpler case where we want to count pairs $n$ and $n+2$ where $n$ is a $2$-almost prime and $n+2$ is prime.
First, for a small prime $p$, we want to find an upper bound for the number of $n < x$ such that $n$ is divisible by $p$ and both $n/p$ and $n+2$ are prime. This is a similar problem to counting twin primes, and in a similar way one obtains a bound of the form $O(x/\log x)$ (key point: the implied constant does not depend on $p$). If we wish to bound the number of pairs $(n,n+2)$ such that $n+2$ is prime and $n$ is a $2$-almost prime, we may instead count the triples $(p,n,n+2)$ where
$p < x$ is prime, $n < x$ is divisible by $p$, $n/p$ is prime, and $n+2$ is prime. If for each $p < x$ we have a upper bound of $Ax/\log x$ (for the same $A$), in total we obtain the upper bound:
$$ \frac{Ax}{\log x} \cdot \sum_{p < x} \frac{1}{p} \sim \frac{Ax \log \log x}{\log x}.$$
Of course, the devil is in the details! END EDIT)
All one needs to answer 3) is that
the exponent of $\log(x)$ in the denominator is $> 1$.
3). Assuming the expected result (*), the inverse sum of the $a$-asterios primes converges.
Consider the set of integers $S_a$
which do not have any prime factors of the form
$\pm 1 \mod a$. This is a reasonable thing to do whenever
$|(\Z/a\Z)^{\times}| > 2$. This is a weaker condition, so there are more of these numbers
and consequently obtaining upper bounds is harder.
We may form the Dirichlet series
$$L(s) = \sum_{S_a} \frac{1}{n^s}$$
which has an Euler product:
$$L(s) = \prod_{p \not\equiv \pm 1} \left(1 - \frac{1}{p^s}\right)^{-1}$$
Now let $K = \Q(\zeta_a)^{+}$ be the totally real subfield of $\Q(\zeta_a)$.
It has degree $r/2 = \phi(a)/2$, where $r > 2$ unless $a = 1,2,3,4$ or $6$.
(What we say now only makes sense for $r \ge 2$, in which case $r/2 \in \Z$.)
A prime splits completely in $K$ if and only if it is of the form
$\pm 1 \mod a$. Looking at the Euler product of $\zeta_K(s)$, we see that,
up to a constant
which can be explicitly written as some product over primes,
$$\zeta_K(s) L(s)^{r/2} \sim \zeta_{\Q}(s)^{r/2}$$
as $s \rightarrow 1^{+}$, and hence $L(s) \sim (s-1)^{(2-r)/r}$
(up to some constant)
as $s \rightarrow 1^{+}$. We deduce (Perron's formula) that the number
of integers $\le x$
all of whose prime factors are not of the form $\pm 1 \mod a$ is
asymptotic to
$$ \kappa \cdot \frac{x}{(\log x)^{2/r}},$$
for some non-zero
constant $\kappa$. This is the same analysis that gives the asymptotic
formula for the number of integers $\le x$ which can be written as a sum
of two squares (a result of Landau).
We immediately deduce:
4a) The number of integers $a$ such that $ak+1$ (or $ak-1$) has no
prime factors of the form $\pm 1 \mod a$ has zero density.
If $r > 2$ (so $a \ge 3$ and $a \ne 3,4,6$) then the power $2/r$ of
$(\log x)$ is at most $1/2$. Thus, we actually
are led to the following guess:
4b) If $r = \phi(a) > 2$, then one would heuristically expect the inverse
sum of integers $k$ such that none of the prime factors of $ak-1$ and $ak+1$
are $\pm 1 \mod a$ diverges. If $r = 2$, so $a = 3$, $4$, or $6$, then (Brun) the series converges.
Here is a related problem of the very same kind: can one count the number of
integers $n \le x$ such that both $n$ and $n+1$ can be expressed as the sum of
two squares, and prove that there are $\sim x/\log(x)$ such integers
(perhaps up to non-zero constant factors)?
Any integer $n$ can be written as the product of two numbers not
of the form $\pm 1$ unless every prime factor of $n$ is of the
form $\pm 1 \mod a$. The integers all of whose prime factors are
$\pm 1 \mod a$ can be analyzed exactly as in the last paragraph.
In this case, the number of integers all of whose prime factors are of the
form $\pm 1 \mod a$ is asymptotic to
$$ \kappa \cdot \frac{x}{(\log x)^{(1-2/r)}}.$$
Suppose that $r > 2$. Then the set of such integers has density zero, and thus the
set of integers which have a factor not of the form $\pm 1 \mod a$
has density one. Any set of density one has infinitely many
"twins" satisfying any fixed congruence condition.
Hence:
5) If $r = \phi(a) > 2$, then there are infinitely many $k$ such that both
$ak+1$ and $ak-1$ have a factor not of the form $\pm 1 \mod a$. Indeed, such numbers have density one.
Finally, I have nothing to say about problem 1) besides the remarks I made in my rephrasing of the original question here:
Chen's Theorem with congruence conditions.
Best Answer
Here is a partial answer to your question. Basically, the methods of Goldston, Graham, Pintz, and Yildirim (see here) have some bearing on your question.
Look at the series
$$\sum_{N < n < 2N} \bigg( \chi_1(n) + \chi_1(n + 2) + \chi_2(n + 2) \bigg) \bigg( \sum_d \lambda_d \bigg)^2$$
where the $\lambda_d$ are the usual Selberg sieve coefficients or some variant, and $\chi_1$ and $\chi_2$ are the characteristic functions of products of exactly one or two primes. Then I worked out a back of a napkin calculation using Theorems 7-9 in GGPY (possibly I made some mistake), and on the Elliott-Halberstam conjecture (a big assumption!) the above is positive, which gives another proof of Chen's theorem (this one conditional).
One nice feature of the GGPY sieve is that it is compatible with the kind of conditions you describe. For example, let $\chi_2$ be the characteristic function of $E_2$ numbers whose prime factors are both 1 mod 4, or split completely in your favorite Galois extension, or some other appropriate condition. (See the GGPY paper as well as a followup which I wrote.) Then the machinery still works. You just multiply the contribution from the $E_2$ numbers by a fourth. In fact you can do better and only sieve on integers $n$ congruent to 1 mod 4, in which case you only have to divide by a half.
According to my napkin calculation, this is numerically not enough, the series is asymptotically slightly negative, and so this argument fails to prove your assertion, even on EH. So perhaps your question really is "difficult" in the way suggested by the parity problem. But this at least shows you how you can hope to prove statements related to what you asked for.