Constructing Turing Machine for division with 8
A number (in binary) is divisible by 8 if it ends at least by 3 0's. Remember that an integer is divisible by 2 (without remainder) if it ends with 0 and since $8=2^3$ its binary representation must end with at least 3 zeros.

In fact a finite automaton is enough to recognize numbers divisible by 8. The following transitions are for a DFA which can be easily rewritten as TM instructions.

$q_1$ - start state, $q_4$ - accept state.
$\delta(q_1, 0) = q_2, \delta(q_1, 1) = q_1$
$\delta(q_2, 0) = q_3, \delta(q_2, 1) = q_1$
$\delta(q_3, 0) = q_4, \delta(q_3, 1) = q_1$
$\delta(q_4, 0) = q_4, \delta(q_4, 1) = q_1$