a) Assume that $a = x + y$, $b = x^2 + y^2$ and $c = x^4 + y^4$ are integers. Then the numbers $a^2 - b = 2xy$ and $b^2 - c = 2x^2 y^2$ are integers as well.

Suppose that $xy$ is not an integer. Then $xy = \frac{m}{2}$, where $m \in \mathbb{Z}$ is odd. But, then $2x^2 y^2 = \frac{m^2}{2}$ is not an integer. Contradiction! Hence, $xy$ is an integer.

By induction we can prove that $x^n + y^n$ is an integer for each $n \in \mathbb{N}$:

Basis of induction: numbers $x + y$ and $x^2 + y^2$ are integers, by assumption.

Let's suppose that, for some integer $n > 1$, the numbers $x^{n-1} + y^{n-1}$ and $x^n + y^n$ are integers.

Then
$$\begin{aligned} x^{n+1} + y^{n+1} &= (x^n + y^n)(x + y) - (x^n y + x y^n) \\ &= (x^n + y^n)(x + y) - xy(x^{n-1} + y^{n-1}). \end{aligned}$$
Numbers $x + y$ and $xy$ are integers, as well as the numbers $x^{n-1} + y^{n-1}$ and $x^n + y^n$, so, we conclude that the number $x^{n+1} + y^{n+1}$ is an integer, too.

b) $x = \sqrt{2}$, $y = -\sqrt{2}$.

c) $x = \frac{1}{\sqrt{2}}$, $y = -\frac{1}{\sqrt{2}}$.