a) Let $a,b,c$ the numbers on the external circle and $a_1$, $a_2$, $a_3$...$a_6$ be the $6$ numbers on the interior circles. a)We have: $a_1$, $a_2$, $a_3$ and $a_4$ are in the same circle. $a_4$, $a_5$, $a_2$ and $a_6$ are in the same circle $a_1$, $a_5$, $a_3$ and $a_6$ are in the same circle. So the following equations are valid: $a+b+c=$ $a_1$+$a_2$+$a_3$+$a_4$ $a+b+c=$ $a_1$+$a_5$+$a_3$+$a_6$ $a+b+c=$ $a_4$+$a_5$+$a_2$+$a_6$, which means $a_1$+$a_2$+$a_3$+$a_4$$=$$a_1$+$a_5$+$a_3$+$a_6$ $\implies$ $a_2$+ $a_4$$=$$a_5$+$a_6$ and we also obtain $a_4$+$a_2$$=$$a_1$+$a_3$ and $a_1$+$a_3$$=$$a_5$+$a_6$ $\implies$ $a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$. So using the new equality in one of the expressions in the beginning we will have $a+b+c=2$ ($a_1$+$a_3$), which means $a+b+c$ is even so we only have two options for the parity of $a,b$ and $c$ which are $2$ odds and $1$ even or all of them even. WLOG let $b$ and $c$ be odd and $a$ even, with this we can have $a=2$ $\implies$ $b+c=2$ ($a_1$+$a_3$$-1$) and here you get the values of the 4 variebles easily but we must remember that $a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$ so with a little bit of casework when we let $b=7$ and $c=9$ $\implies$ $a_1$+$a_3$$=$ $\frac{7+9+1}{2}$$=$$9$ therefore $a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$$=$$9$ so we can have $a_1$$=$ $1$ $a_3$$=$ $8$ $a_2$$=$ $5$ $a_4$$=$ $4$ $a_5$$=$ $3$ $a_6$$=$ $6$. And checking $7+9+2$$ =$ $1+5+8+4$$=$ $1+6+3+8$$=$ $4+3+5+6$, we see that it works so its a solution.

b) For the sake of a contradiction lets suppose there is a solution, so from a) we obtained that $a+b+c$ is even and we only have 2 options. If all of them are even then $a,b,c$ can only be $2,4,6$ or $8$ then the only possible triple we can have are $(2,4,6)$ , $(2,4,8)$ and $(4,6,8)$. Case $(2,4,6)$ The $a+b+c=12$ $\implies$ $a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$$=$$6$ but notice that we know that $9$ must be one of the numbers in the internal circles which doesn't make sense because the other number that's paired with him has to be $-3$. Case $(4,6,8)$ Then $a+b+c$$=$$18$ $\implies$ $a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$$=$ $9$ and again $9$ is one of the numbers in the internal circles so the other number that's paired with must be $0$, absurd. Case $(2,4,8)$ Then $a+b+c=14$ $\implies$ $a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$$=$ $7$ and by the same argument as the two previous cases we get a contradiction. With this we have that if there's a solution it must be with $2$ odds and $1$ even so WLOG let's suppose $a_6$$=$ $9$, $a$ be even and $b,c$ odds. Now by the equality we obtained in a) $2$($9$+$a_5$)=$a+b+c$ now I claim that $a_5$ is even because since $a$ is even and $b,c$, $a_6$ are odd we'll only have $2$ odds and $3$ even numbers left but we know that$a_1$+$a_3$$=$ $a_2$+$a_4$$=$$a_5$+$a_6$ and the only chance for the parity to be equal is if all the sums are odd then each pair must have $1$ odd and $1$ even and since $a_6$ is odd then $a_5$ must be even. Now we noticed that $18$+$2$$a_5$$\geq 18+(2)(2)=22$ but $a+b+c$$\le 7+5+8=20$ so there's no solutions. $\square$.