a) Notice that $x = 1, y = 3, z = 8, t = 4$ is a solution to the system $(S_7)$, and so is $(k, 3k, 8k, 4k)$, for any $k \in \mathbb{N}$, hence $7 \in M_1$.

If $(x, y, z, t)$ is a solution to the system $(S_{10})$, it would follow that $11(x^2 + y^2) = z^2 + t^2$. From $11 \mid z^2 + t^2$ we get $11 \mid z$ and $11 \mid t$. Then $11 \mid x^2 + y^2$, hence $11 \mid x$ and $11 \mid y$. Therefore, there exist $x_1, y_1, z_1, t_1 \in \mathbb{N}$ such that $x = 11x_1, y = 11y_1, z = 11z_1, t = 11t_1$. This leads to $11(x_1^2 + y_1^2) = z_1^2 + t_1^2$. By continuing this procedure ("infinite descent"), we get that $x$ is a multiple of any power of $11$, and, as $x \neq 0$, we arrive to a contradiction. In conclusion, the system $(S_{10})$ has no solutions, i.e. $10 \in M_2$.

b) It is easy to see that any number $n$ of the form $m^2 - 1$, $m \in \mathbb{N}$, belongs to $M_1$: we can choose $x = y$ and notice that $(k, k, mk, mk)$, $k \in \mathbb{N}$, are solutions to the system $(S_{m^2-1})$, hence $m^2 - 1 \in M_1, \forall m \in \mathbb{N}$. Applying the same steps as we did above for $n = 10$, it is easy to prove that if $p$ is a prime of the form $4m + 3$, and $n = p - 1$, then $n \in M_2$. As there are infinitely many such primes, the conclusion follows readily. One can prove that actually any number congruent to $2 \pmod{4}$ belongs to $M_2$.