Since the OP asked for other examples of this kind of numerology,I will give another one to support his observation
The function $\cos(\theta_{11})$ has the following closed form
$\cos(\theta_{11})=\frac{\sigma_{1}(11)}{22\sqrt{11}}-\frac{12\sum_{k=1}^{10} (2178k^2-572k^3+35k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag1$
and continued fraction
$\cos(\theta_{11})=\{0;1,1,572,3,2,1,2,1,2,2,4,3,1,6,\dots\}\tag2$
where we clearly see $572$ appearing both as a coefficient in the sum $(1)$ and largest partial quotient in the first few partial quotients of the continued fraction $(2)$
Is this a coincidence?
Edited:03 Sep 2017
And also
$2\cos(\theta_{11})=\frac{\sigma_{1}(11)}{11\sqrt{11}}-\frac{12\sum_{k=1}^{10} (4356k^2-1144k^3+70k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag3$
with the following continued fraction
$2\cos(\theta_{11})=\{1;1145,1,1,2,6,2,2,1,1,1,1,1,2,3,\dots\}\tag4$
where $1145$ is the 0th partial qotient in the continued fraction $(4)$ and $1144$ appears as a coefficient in the formula $(3)$
Remark: Identity $(1)$ is a special case of the identity found in A000594 OEIS when $n=11$
$\tau(n)=n^4\sigma_{1}(n)-24\sum_{k=1}^{n-1} (18n^2k^2-52nk^3+35k^4)\sigma_{1}(n-k)\sigma_{1}(k)$
I can answer your first question. In arXiv:1208.4074 by Dabholkar, Murthy and Zagier you can find a formula that implies
$H^{(2)}(\tau)= \frac{48 F_2^{(2)}(\tau)- 2 E_2(\tau)}{\eta(\tau)^3}$
where $E_2(\tau)$ is the quasi modular Eisenstein series and
$F_2^{(2)}(\tau)= \sum_{r>s>0,r-s\ \mathrm{odd}} (-1)^r s\, q^{rs/2}$ which you can use to compute $H^{(2)}(\tau)$ to whatever order you desire. I have no idea concerning your questions 2,3, and 4.
Best Answer
Yes, this is true, as a consequence of an identity in a space of modular forms of weight $6$.
The form $\eta(2z)^{12}$ is in this space; and $\sigma_5(n)$ for $n$ odd are the coefficients of the weight-$6$ form $$ \frac1{1008} \Bigl(E_6(z+\frac12) - E_6(z)\Bigr) = q + 244 q^3 + 3126 q^5 + 16808 q^7 + 59293 q^9 + \cdots. $$ The difference is $$ 256(q^3 + 12 q^5 + 66 q^7 + 232 q^9 + 627 q^{11} + 1452 q^{13} + \cdots); $$ after a bit of experimentation we recognize this as $256q^3$ times the $12$-th power of $1+q^2+q^6+q^{12}+q^{20}+\cdots$, which is to say $256$ times the $12$-th power of the weight-$1/2$ modular form $\sum_{k=0}^\infty q^{(2k+1)^2/4}$. Such a formula, once surmised, can be proved by comparing initial segments of the $q$-expansions; I did this to $O(q^{61})$, which is way more than enough. The congruence mod $256$ follows because the $12$-th power of $1+q^2+q^6+q^{12}+q^{20}+\cdots$ clearly has integral coefficients.
Added later: This identity (and thus the congruence that Tito Piezas III asked for) gives a formula $(\sigma_5(n) - \rho(n)) / 256$ for the number of representations of $4n$ as the sum of $12$ odd squares, or equivalently of $(n-3)/2$ as the sum of $12$ triangular numbers. Following this lead, I soon found both the formula and the congruence in the paper
available on Ken Ono's page, where the enumeration is given (in the triangular-number form) as Theorem 7, and the congruence as a "simple consequence" of that theorem.