a) The only 4-tuples satisfying the required conditions are
$$(1, 2, -1, -2) \quad \text{and} \quad (1, -2, -1, 2)$$
along with their cyclic permutations; the cyclic sum in question is $-3$ for the former and its cyclic permutations, and $3$ for the latter and its cyclic permutations.
The condition that the cyclic sum of the $x_k/x_{k+1}$ be integral is equivalent to the cyclic sum of the $x_k^2 x_{k+2} x_{k+3}$ being of the form $N x_0 x_1 x_2 x_3$ for some integral $N$. Since the $x_k$ are pairwise distinct, and each $x_k$ is coprime to $x_{k+1}$, it follows that $x_2 = -x_0$ and $x_3 = -x_1$, so $2(x_0^2 - x_1^2) = N x_0 x_1$. Since $x_0$ and $x_1$ are relatively prime, so are $x_0^2 - x_1^2$ and $x_0 x_1$, hence the latter is a divisor of $2$. The required 4-tuples are now easily derived.

b) If $k$ is an integer greater than $1$, then $(1, -k, k+1, -1, -k^2 - k)$ is a 5-tuple satisfying the required conditions; the corresponding cyclic sum is $-k^2 - 2k - 2$.