Solution:

The answer to part (b) is yes.
For brevity, we use the notation of complex numbers. For any integers $x$ and $y$, not both zero, let $z = x + y i$ and
$$P(z) = \frac{z^{2}}{\bar{z}^{2}}$$
Clearly, $|P(z)| = 1$ so $P(z)$ is a point on the unit circle. We claim that for any $z_{1} = x_{1} + y_{1}$ and $z_{2} = x_{2} + y_{2}$, the distance between $P(z_{1})$ and $P(z_{2})$ is rational. We compute:
$$\begin{aligned} \left|P(z_{1}) - P(z_{2})\right|^{2} & = \left(P(z_{1}) - P(z_{2})\right)\left(\overline{P(z_{1})} - \overline{P(z_{2})}\right) \\ & = \left(\frac{z_{1}^{2}}{\bar{z}_{1}^{2}} - \frac{z_{2}^{2}}{\bar{z}_{2}^{2}}\right)\left(\frac{\bar{z}_{1}^{2}}{z_{1}^{2}} - \frac{\bar{z}_{2}^{2}}{z_{2}^{2}}\right) \\ & = -\frac{z_{1}^{2} \bar{z}_{2}^{2}}{\bar{z}_{1}^{2} z_{2}^{2}} + 2 - \frac{\bar{z}_{1}^{2} z_{2}^{2}}{z_{1}^{2} \bar{z}_{2}^{2}} \\ & = -\left(\frac{z_{1} \bar{z}_{2}}{\bar{z}_{1} z_{2}} - \frac{\bar{z}_{1} z_{2}}{z_{1} \bar{z}_{2}}\right)^{2} \end{aligned}$$
The expression in parentheses is purely imaginary (being the difference of a complex number and its conjugate). Its imaginary part is rational since the components of $z_{1}$ and $z_{2}$ are rational. We conclude that $\left|P(z_{1}) - P(z_{2})\right|$ is rational.
To finish the proof, it suffices to show that $P(z)$ takes on infinitely many values. It is not hard to check that the choices $z = 1 + i, 1 + 2i, 1 + 3i, \ldots$ all give distinct values of $P(z)$.