a) Points of the form $(q,\alpha)$ or $(\alpha, q)$, where $q\in\mathbb{Q},\alpha\in\mathbb{R}\setminus\mathbb{Q}$, cannot lie on a line passing through two rational points, since solving the equation of the line for $\alpha$ would yield a rational value. Hence (a) is false. b) If the given point is of the form $(q_1, q_2)$, where $q_1,q_2\in\mathbb{Q}$, then take $(q_1\sqrt{2},q_2\sqrt{2})$ and $(2q_1\sqrt{2},2q_2\sqrt{2})$ as the desired points. If the given point is of the form $(\alpha_1,\alpha_2)$, where $\alpha_1,\alpha_2\in\mathbb{R}\setminus\mathbb{Q}$, then take $(2\alpha_1,2\alpha_2)$ and $(3\alpha_1,3\alpha_2)$ as the desired points. If the given point is of the form $(q,\alpha)$, where $q\in\mathbb{Q},\alpha\in\mathbb{R}\setminus\mathbb{Q}$, then: 1. If $\alpha$ is an algebraic number, take $(q\pi,\alpha\pi)$ and $(2q\pi,2\alpha\pi)$ as the desired points; 2. If $\alpha$ is a transcendental number, take $(q\sqrt{2},\alpha\sqrt{2})$ and $(2q\sqrt{2},2\alpha\sqrt{2})$ as the desired points. Similarly if the given point is of the form $(\alpha, q)$. Therefore (b) is true.

Part (b) has this more elegant solution: Choose an arbitrary line through the given point, not parallel to one of the coordinate axes. It suffices to show that there are at least two irrational points on the line. Consider $\pi_1, \pi_2$ the parallel projections of the line onto the coordinate axes, these are bijections. Now consider the preimages of the rational numbers under both of these projections. These are countable subsets of the line, thus they can't cover it completely, there are uncountably many points left. These have to be irrational points.