Let $A+B+C+D=T$. Then, $5T=A+B+C+D+E+F+G+H+2(A+B+C+D)=36+2T\Rightarrow T=12$. Hence, the sum of the numbers on each of the five circles is $12$. $A,B,C,D\in [1,8]$ and $A+B+C+D=12$. Thus, $\{A,B,C,D\}=\{1,2,4,5\}$ or $\{A,B,C,D\}=\{1,2,3,6\}$. i) $\{A,B,C,D\}=\{1,2,4,5\}$ One of the numbers $E,F,G,H$ is $8$. Let this number be $E$. Then, $A+B=4$ and $\{A,B\}=\{1,2,4,5\}$. No solution here. ii) $\{A,B,C,D\}=\{1,2,3,6\}$ One of the numbers $E,F,G,H$ is $8$. Let this number be $E$. (We have $4$ options here). Then, $A+B=4$ and $\{A,B\}=\{1,2,3,6\}$. Hence, $\{A,B\}=\{1,3\}$. WLOG $A=1$ and $B=3$. (We have $2$ options here). If $D=2$, then $H=9$ Contradiction. Thus, $D=6$ and $C=2$. From here, we can find that $F=7, G=4, H=5$. So, when we determine which number is $8$ and the order of its neighbors, we have a unique distribution for the others. Thus, we have $4\times 2=8$ solutions in total.