Prove that if $k$ is highly abundant or highly composite and $q,p$ are the nearest primes with $q+1<k<p-1$, then $k-q,p-k$ are primes. This immediately implies that all highly abundant and highly composite numbers are the sum of two primes and the difference of two primes.
I've checked this for the first $1231$ highly abundant numbers and the first $98$ highly composite numbers
Update: I've found this weak form of the conjecture, which has been checked much further.
Update 2: This is related to several conjectures other than the one linked. Most notably though it's a "more general" version of Fortune's conjecture (in that the highly composite and highly abundant numbers grow much more slowly than primorials). Another conjecture of this general form (that the least $a,b>1$ such that $f(n)+a,f(n)-b$ are prime are themselves prime) is made for factorials. I would appreciate references to any other conjectures of the same form. The values $k-q$ and $p-k$ from above are analogous to the lesser and greater Fortunate numbers respectively.
Take your letter $k$ to stand for a primorial number, $k = 2 \cdot 3 \cdots p_r,$ the product of consectutive primes up to some $p_r.$ Now, by the prime number theorem, $k$ is approximately $e^{p_r}.$ In turn, the expected gap between primes at for that size of numbers is about $p_r,$ and numerical experiments say that the largest gap is $p_r^2,$ although no proof whatsoever for this, it is called Cramer's conjecture. Since your $$ |k-q| < p_r^2, \; \; |k-p| < p_r^2, $$ and both $k-p$ and $k-q$ fail to be divisible by any prime up to and including, it follows that both differences are prime, essentially by an "effective" version of Cramer's conjecture.
However, and this will require some huge numbers, you were to learn how to program the superior highly composite numbers and the colossally abundant numbers, both these types of numbers are, eventually, much much larger than $e^{p_r},$ where $p_r$ is the largest prime dividing your $k.$ Even with Cramer's conjecture, the odds are that your differences $|k-p|$ and $|k-q|$ are, eventually, likely to be composite. To program this, it will be necessary to implement partial factoring, to rule out composite numbers by small trial factors.
An easier experiment is this: both the SHCN and the CAN are "approximately" $$ k = \operatorname{LCM} \{ 1,2,3,4, \ldots, n-1,n \} $$ for some $n.$ So, my recommendation is to program your test for $k$ the least common multiple of the numbers from $1$ to some $n.$ Then do the best you can casting out nearby composite numbers; find some nearby probable primes; confirm them with, say, PRIMO. Then factor your differences. It will be a good deal of work, but worth understanding the problem.
Ummm. In short, your conjectures, as written, are unlikely in the long run. Probably true for the primorials, though.
EDIT: the exponent of $p$ in $\operatorname{lcm}(1,2,3,\ldots,n)$ is $$\left\lfloor \frac{\log n}{\log p} \right\rfloor$$ So, for two primes that are fairly small in comparison to $n,$ the ratio of the exponent of $p$ over exponent of $q$ is approximately $$ \frac{\log q}{\log p}. $$
Now, in the S.H.C. numbers and the colossally abundant numbers, in fact for all Ramanujan's "Generalised Superior Highly Composite Numbers," For a small real $\delta > 0,$ the ratio of the exponent of $p$ over exponent of $q$ is approximately $$ \frac{q^\delta - 1}{p^\delta - 1}. $$ By L'Hospital's Rule, in the limit as $\delta$ goes to $0^+,$ this approaches $ \frac{\log q}{\log p}. $
Ramanujan took real parameter $s,$ it suffices for our purposes to take $s > 0.$ Then the GSHCN corresponding to $s$ is specified by displaying the exponent on each prime factor $p.$ And, from formula (311), the exponent is $$ \color{magenta}{a_p = \left\lfloor \frac{\log \left( \frac{p^\delta - p^{-s}}{p^\delta - 1} \right)}{s \log p} \right\rfloor}$$ The colossally abundant numbers of Alaoglu and Erdos are the case $s=1.$ To get the original Superior Highly Composite numbers of Ramanujan,, we fix $\delta$ and take the limit as $s \rightarrow 0^+,$ resulting in $$ \color{magenta}{ a_p = \left\lfloor\frac{ 1}{p^\delta - 1}\right\rfloor}. $$