All right, the calcium atom, when ionized, loses two electrons; specifically, it loses the two electrons in its valence energy level. In the case of calcium, its valence energy level is its fourth energy level. With the fourth energy level depleted, the third energy level is exposed, whose radius is smaller (naturally) than the fourth energy level. Hence, the radius of the Ca2+ ion is understandably smaller than the radius of the Ca atom.
Potassium has 19 protons and 19 electrons, while calcium has 20 of each. As a result, there is a greater electrostatic attraction between a Ca nucleus and its electrons than there is between a K nucleus and its electrons. The valence electron in a potassium atom therefore sits further away from the nucleus and is more easily plucked off (which is to say, it requires less energy to be removed from the atom). Calcium's first valence electron is slightly more difficult to remove because it is closer to the positively charged nucleus.
The second ionization energy of potassium is a whole lot larger than the first ionization energy. This is because the removal of the first electron from a potassium atom leaves behind (A) a positive ion and (B) an ion whose valence energy level has a noble gas configuration ([Ar]). Because this configuration is very stable, removing a second electron from a K+ ion requires an enormous expenditure of energy. K2+ ions could only exist in very high-energy environments.
The second ionization energy of calcium IS higher than its first ionization energy. You are, after all, removing an electron from a positively charged ion; however, the Ca+ ion does not have a noble gas configuration, so it easily loses a second electron to achieve one. If you were to look up calcium's THIRD ionization energy, you would find it to be tremendously higher than the second, just as potassium's second ionization energy is much higher than its first.
To ionize magnesium once (Mg --> Mg+), you have to remove an electron from a complete s-sublevel, a fairly stable environment. The Mg+ ion is thermodynamically less stable than either the Mg atom or the Mg2+ ion, which is why you rarely see it in nature. The first electron removed from aluminum, however, comes not from a complete s-sublevel, but from a half-filled p-orbital. It shouldn't take as much energy to remove that first electron since doing so doesn't disrupt a fairly stable electron arrangement. Removing the SECOND electron from Al+, however, WILL form a half-empty s-orbital. As a result, you would expect aluminum's second ionization energy to be considerably larger than it's first (and not just because it involves ionizing a cation).
I know that's more than you needed, but my goal is to help you understand rather than just doing your homework for you. I hope it helps and good luck.