You may have forgotten what the notation [Xe]6s24f145d106p1 means. Here we just need to know the number of electrons, and thereby the atomic number, to identify the element. The [Xe] part is a shorthand to indicate that we start with the structure of Xenon, and then add the "6s24f145d106p1" part. Xenon is 54 electrons (from the periodic table in the Databook) and in the remainder of the notation it is the superscripts that are the numbers of electrons -- so we sum 2+14+10+1 = 27 and then 54 + 27 = 71. Thallium (Tl) is the element with atomic number 71. The trap here is that you may think that "6s" and "4f" refer to numbers of electrons. They don't; they are the names of atomic orbitals. The best reference I see to the material in these questions is section 6.9 of Chembook (p.204).
Continuing from the previous analysis, here we start with the number of electrons for [Ar] (18), then add 5 (from 3d5), and now an additional two since a 2+ ion is missing 2 electrons. The result is the element with atomic number 25 -- manganese. Note that they really did ask for the symbol of the ion (Mn2+), not the element.
Well you could be forgiven (by me, not IB) if you did not guess that ns2np3 means any of 1s21p3, 2s22p3, 3s23p3, etc. (It could be some symbolism you have half forgotten about "hybridized orbitals" or something, but it is not.) Once you get there you should be saying to yourself that the atomic number, equal to the number of electrons, must be that of a noble gas plus 5 (the sum of the superscripts in ns2np3). So He + 5 = N, Ne + 5 = P, etc. (The fact that they say "elements with a ground state of ... " indicates we are not talking about ions.)
Another way you might have solved this problem is by remembering that the 'p' orbitals contain 6 electrons when full -- with the 2 's' electrons you get those figures of molecules with 8 dots -- so, since we start here with only 3, we are looking for the column of elements that form 3 bonds. The one you should always remember is Nitrogen, as in NH3 (ammonia), and P (phosphorus) is just below N in the periodic table.
Obviously, if you had the right sort of diagram, you would have 2 of three points in your pocket. But I fail to find that diagram in either of the chemistry books - the only decent example is in Physicsbook fig. 27-28 pg. 843, though it would be nicer if the x axis was frequency rather than wavelength so that it went up, rather than down, in energy.
Briefly, hydrogen, with its one electron, has the simplest of "emission spectrums" of any element, and the only one we can calculate exactly from theory. Light is emitted when the electron falls from one orbital to another of lower energy, and the frequency of the light emitted corresponds exactly to the energy difference - symbolically ΔE = hf (as used in IBbook, your Databook and Physicsbook or ΔE = hν as in Chembook (pg. 178), but not ΔE = hυ as in IB's answer.
The simplification due to the fact that there is only one electron is that the s, p, d, f orbitals in a given shell are all the same energy, unlike the situation for any other element. This means that there are enormously fewer energy differences, and therefore fewer spectral lines, than in any other element's spectra. (So you are not likely to be asked this question about any other element.)
A key concept required to get the full credit here is that the spectral lines get closer and closer together as you go to higher energies (higher frequency, shorter wavelength). The keyword "converge" is worth a whole point here. Why?
Well, as one goes from lowest to higher (energy) orbitals, the energy gap between orbitals decreases; it disappears entirely (converges) with an essentially infinite number of levels just below a particular energy, and that is the so-called "ionization" energy. This energy is that required to rip an electron entirely off of the element. Now spectra are energy jumps between orbitals, and all possible transitions could result in spectral lines of the appropriate energy. But really we only see (even in the sun) the jumps that result from the falling of electrons from high energy orbitals down into the 2 or 3 lowest orbitals because of population probabilities. This means that there are a family of lines that show the drop from high into the lowest (1s) orbital, another family from drops into the 2nd (2s or 2p) orbitals, etc. In these families, the highest energy jump is from the ionization energy into the relevant orbital, and then there are all the drops from the nearly infinite number of orbitals near ionization, and then more space as we go down to the orbital right above the one we are dropping into. For a reference section, try Chembook fig. 6.13 and section 6.3 (pg. 181).
It should be easy enough to say that if you take something away (an electron) then what is left should be smaller. IB requires that you note that for Li, the electron lost is the only one in the outer (#2) shell. (In fact whenever we are talking about the ions of elements in the first two columns of the periodic table, we are talking about elements happy to shed the starting electrons of a new shell to get back to the "noble gas" state of a complete inner shell.)You must also know that removing an electron makes all the other electrons more lonely -- they push each other away from the nucleus, so removal of one allows the others to draw closer. The technobabble way of saying this is to say that the "effective charge" of the nucleus is larger for the ion relative to the neutral atom.
IB's answer -- given any question about the noble (or "inert") gases -- is to say everything they know about them. This would seem to be a question to find out if you know what electronegativity is, but the answer given is vague and lame. Chembook has the explicit definition "Electronegativity is defined as the ability of an atom in a molecule to attract electrons to itself." So the fact that the noble gases do not form molecules lets them out of this 'electronegativity' thing altogether.