(a), (b) It is better to think about the space as described in the solution for the exercise 1: we just add a point and new basis neighborhoods of the point. The intersection of a new basis neighborhood with an old one is either an interval or the union of two intervals or empty. Also, this way it is immediate that the space without the new point is homeomorphic to the real line, and if we substitute the new point for 0, then it is also homeomorphic to the set of real numbers.(c) It satisfies the $T_1$ axiom: in fact, any pair of points except $(p,q)$ can be separated by two neighborhoods, while these two points cannot be separated by two disjoint neighborhoods, though each one has a neighborhood not containing the other one.(d) $X-\{p\}$ and $X-\{q\}$ are both open, metrizable, and each point belongs to at least one of these open sets. Also, since $\mathbb{R}$ is second-countable, so is $X$ , as $(-q,q)$ where $q\in\mathbb{Q}$ is a basis at the new point.