Let $O$ be the origin of coordinates. Let $A_0$ be the starting point of the beetle's journey, $A_1$ the first turning point, $A_2$ the second turning point, $A_3$ the third turning point and $A_4$ the endpoint (see figure below).

The right triangles $OA_0A_1$ and $OA_1A_2$ are similar with ratio $x$ because $\angle OA_1A_2 = 90^\circ - \angle OA_1A_0 = \angle OA_0A_1$ and $\frac{OA_1}{OA_0} = x$. So are the triangles $OA_1A_2$ and $OA_2A_3$ similar with ratio $x$ since $\frac{OA_2}{OA_1} = x$ by similarity of the triangles $OA_0A_1$ and $OA_1A_2$. Analogously, the triangles $OA_2A_3$ and $OA_3A_4$ are similar with ratio $x$. Consequently, $OA_2 = x^2$, $OA_3 = x^3$ and $OA_4 = x^4$.

By the Pythagorean theorem, the length of the beetle's journey is $\sqrt{1+x^2} + \sqrt{x^2+x^4} + \sqrt{x^4+x^6} + \sqrt{x^6+x^8}$, or equivalently, $(1+x+x^2+x^3)\sqrt{1+x^2}$ which equals $(x^4-1) \cdot \frac{\sqrt{x^2+1}}{x-1}$ if $x \neq 1$. On the other hand, the distance between $A_0$ and $A_4$ is $|x^4-1|$.

a) Taking $x = \frac{4}{3}$, the number $x^4 - 1$ is rational, since $x$ is rational. As $\sqrt{1+x^2} = \sqrt{1+\frac{16}{9}} = \frac{5}{3}$, the number $(x^4-1) \cdot \frac{\sqrt{x^2+1}}{x-1}$ is also rational.

b) Assume that the distance between $A_0$ and $A_4$ is an integer; then $x^4$ is an integer. Suppose that $(x^4-1) \cdot \frac{\sqrt{x^2+1}}{x-1}$, the length of the beetle's journey, is also an integer (one can assume $x \neq 1$ since otherwise the length of the journey is $4\sqrt{2}$ that is not an integer). As $x^4 - 1$ is an integer, the number $\frac{\sqrt{x^2+1}}{x-1}$ must be rational. Consider three cases.

1) If $x$ is an integer then $x-1$ is integer whence $\sqrt{x^2+1}$ must be rational and $x^2+1$ must be a perfect square. This is impossible as two consecutive positive integers cannot be perfect squares.

2) Suppose that $x$ is irrational and $x^2$ is an integer. As $\frac{x^2+1}{(x-1)^2}$ is the square of a rational number, so is also $\frac{(x-1)^2}{x^2+1}$. Hence $\frac{2x}{x^2+1}$ must be rational. But this is impossible, since $2x$ is irrational and $x^2+1$ is an integer.

3) Let $x^2$ be irrational. Similarly to the previous case we see that $\frac{2x}{x^2+1}$ is rational. Hence $\frac{4x^2}{(x^2+1)^2}$ is the square of a rational number, implying that $\frac{(x^2+1)^2}{4x^2}$ is rational. The latter in turn implies that $\frac{x^4+1}{4x^2}$ must be rational. This is impossible as $4x^2$ is irrational and $x^4+1$ is a positive integer.