Carbon-13 NMR for chloroform
You assumed that coupling to the three chlorides would yield some type of quartet. This is correct in principle. However, chlorine is one of the many quadrupole nuclei that are basically unobservable by NMR due to their rapid relaxation. I hope another answer is around explaining that as I am not good at it.

I can, however, help you interpret the $1:1:1$ triplet of your spectrum. It is not, as you have assumed, the signal for $\ce{CHCl3}$ but that for $\ce{CDCl3}$. The difference between the two in NMR terms is that chloroform contains the spin ½ nucleus protium while deuterochloroform contains the spin 1 nucleus deuterium — and the latter does not relax rapidly as the chlorine isotopes do. Hence, deuterium is well-observable by NMR.

Now your first thought should be something like the following:


  But I am only coupling to one nucleus. That should give me a $1:1$ doublet and not a $1:1:1$ triplet, shouldn’t it?


That simplification is only correct for spin ½ nuclei. The actual formula for calculating the multiplicity of a peak is $2\,n\,I + 1$, where $n$ is the number of chemically equivalent nuclei your observed nucleus is coupling to and $I$ is the spin of the coupling nuclei. Since deuterium is spin 1 ($I = 1$), the formula gives us:

$$2\,n\,I + 1 = 2 \times 1 \times 1 + 1 = 3$$

Or an expected triplet. Since this triplet does not derive from two equivalent couplings to spin ½ nuclei, it is not a $1:2:1$ triplet but a $1:1:1$ one.



Your second spectrum shows a $\ce{^13C}$ NMR of deuterochloroform which still contains residues of chloroform — at least, that is what it is supposed to show. For some reason unknown to me, the protochloroform signal has not been decoupled, i.e. the spectrum is in fact $\ce{^13C}$ and not $\ce{^13C\{^1H\}}$. Thus, you are observing a strong coupling to the single hydrogen nucleus giving a $1:1$ doublet as expected if chlorine is ignored.