Solution:

a. Let us first observe that $x=1, y=2, z=2$ is a solution, since $1^{2}+2^{2}+2^{2}=3^{2}$. It follows then that, for every positive integer $n$, the triple $x=n, y=2 n, z=2 n$ satisfies the requirement of the problem, since
$$n^{2}+(2 n)^{2}+(2 n)^{2}=9 n^{2}=(3 n)^{2}.$$

b. For every positive integer $t$, the triple $1,2 t, 2 t^{2}$ satisfies the requirements of the problem: indeed the greatest common divisor of these three numbers is always equal to 1, and we have
$$1+(2 t)^{2}+\left(2 t^{2}\right)^{2}=1+4 t^{2}+4 t^{4}=\left(1+2 t^{2}\right)^{2}.$$
A possible way to find this solution is the following: first of all, to ensure that the condition $\operatorname{MCD}(x, y, z)=1$ is satisfied we choose $x=1$. We can then hope that $x^{2}+y^{2}+z^{2}=1+y^{2}+z^{2}$ can be identified with the square of a certain binomial. Writing $1+y^{2}+z^{2}=(a+b)^{2}=a^{2}+2 a b+b^{2}$, it is natural to choose $a=1, 2 a b=y^{2}$ and $b=z$. Solving this system we find $z=b=\frac{y^{2}}{2}$, and clearly in order for $z$ to be an integer it is necessary that $y$ be even, so we write $y=2 t$ for some positive integer $t$: we then find the proposed solution, $x=1, y=2 t, z=\frac{y^{2}}{2}=2 t^{2}$.