For Question $3$ about the recurrence relations, using Mathematica, for $a_n := T_{7A}(n)$ I found:
$$ 0 = 14(n+1)(n+2)(2n+3) a_n \\ -3(n+2)(19n^2+76n+80) a_{n+1} \\ + 5(2n+5)(3n^2+15n+19) a_{n+2} \\ - (n+3)^3 a_{n+3}. $$
For $b_n := T_{7B}(n)$ there are several recurrences. For degree $4$ polynomials I found:
$$ 0 = -7^5(n-14)(n+1)^3 b_n \\ -7^3(19n^4 -1450n^3 -2858n^2 -5586n -3612)b_{n+1} \\ -7(85n^4 -7687n^3 -63795n^2 -173113n -157528)b_{n+2} \\ +(85n^4 +9727n^3 +92931n^2 +311209n +355082)b_{n+3} \\ +(19n^4 +906n^3 +9346n^2 +36306n +48840)b_{n+4} \\ +(n+20)(n+5)^3 b_{n+5}. $$
Alternatively, if $n\to k-3,$
$$ 0 = -7^5(k-17)(k-2)^3 b_{k-3}\quad \\ -7^3(19k^4-678k+2218k^2-2640k+1113)b_{k-2}\;\; \\ -7(85k^4-8707k^3+9978k^2-7072k+2090)b_{k-1} \\ +(85k^4+8707k^3+9978k^2+7072k+2090)b_{k}\, \\ +(19k^4+678k+2218k^2+2640k+1113)b_{k+1} \\ \quad +(k+17)(k+2)^3 b_{k+2}. $$
For degree $3$ polynomials I found:
$$ 0 = 7^7(n+1)^3b_n \\ + 7^5(47n^3 +264n^2 +502n +322)b_{n+1} \\ + 2\cdot 7^3(480n^3 +3930n^2 +10883n +10166)b_{n+2} \\ + 7^2 (1578n^3 +16935n^2 +61249n +74595)b_{n+3} \\ + 7 (1578n^3 +20937n^2 +93265n +139493)b_{n+4} \\ + 2 (480n^3 +7590n^2 +40163n +71138)b_{n+5} \\ + (47n^3 +864n^2 +5302n +10862)b_{n+6} \\ + (n+7)^3 b_{n+7}. $$
Alternatively, if $n\to k-4,$
$$ 0 = 7^7(k-3)^3b_{k-4}\; \\ + 7^5(47k^3-300k^2+646k-470)b_{k-3} \;\\ + 2\cdot 7^3(480k^3-1830k^2+2483k-1206)b_{k-2}\quad \\ + 7^2 (1578k^3-2001k^2+1513k-433)b_{k-1} \\ + 7 (1578k^3+2001k^2+1513k+433)b_{k}\;\; \\ \;\; + 2 (480k^3+1830k^2+2483k+1206)b_{k+1} \\ \; + (47k^3+300k^2+646k+470)b_{k+2} \\ \; + (k+3)^3 b_{k+3}. $$
