Cl and Br have huge quadrupolar moments and their effect to other nuclei in the neighborhood can be considered non-existent.

The effect of 14N is dependent on its electron environment.  3-coordinated 14N has a large asymmetry in electron structure and thus the quadrupolar coupling is large, thus usually has negligible effect to neighboring nuclei.  4-coordinated 14N is more electronically symmetric and has a smaller quadrupolar coupling, therefore can split neighboring nuclei.  In this case, the neighboring 13C will be split into a triplet with intensity ratio of 1:1:1.  The splitting distance gives you the J-coupling constant between the nucleus of interest and the 14N.

2H has a fairly small quadrupolar coupling so it almost always split neighboring 13C to a 1:1:1 triplet.  Those who have run 13C spectra of samples with CDCl3 must be quite familiar with those triplets!

If you use acetone-d6 or dmso-d6 as solvent, the solvent peak on 13C spectra is not triplet but a septulet.  Do you know why and can you predict the intensity ratio of the 7 peaks?

31P and 19F have a spin of 1/2 and will always split neighboring nuclei to a 1:1 doublet.  If multiple 19F are present in the neighborhood, your 13C peaks will have a more complex splitting.