i analyzed the first one manually```
f(n) = 2^2^(n+1)
g(n) = (5*2^(2^(f^n(0)+1)+2)-8)/9
n0 = (5*2^(2^(2^32+1)+1)-4)/9
n1 = 2^(2^32+1)-4
0RB1RZ0RB_1RC1LB2LB_1LB2RD1LC_1RA2RC0LD halts with 5*2^(2^(f^g^n1(n0)(0)+1)+2)+7 nonzero symbols on the tape, after roughly (that)^2 steps
that's around 2^^^(2^(2^32+1)-1)```