Given an 8-digit number $a$ that satisfies the conditions in the problem, let $a_{i}$ denote the difference between its $(i+1)$ th and $i$ th digit. Since $i \in\{1,2\}$ for all $1 \leq i \leq 7$, we have $7 \leq a_{1}+a_{2}+\cdots+a_{7} \leq 14$. The difference between the last digit and the first digit of $m$ is $3-1 \equiv 2$ $(\bmod 10)$, which means $a_{1}+\cdots+a_{7}=12$. Thus, exactly five of the $a_{i} \mathrm{~s}$ equal to 2 and the remaining two equal to 1 . The number of permutations of five 2 s and two 1 s is $\binom{7}{2}=21$.