Protons on aromatic rings in NMR
This is true that all the aromatic protons come near the region having $\delta$: $\pu{7\!-\! 8 ppm}$. But you can always easily find examples where all the aromatic protons doesn't have the same chemical shift (which definitely include your examples). In fact the cases where all the aromatic protons have exactly similar chemical shift is hardly found. 

The main idea is the difference between the terms Chemical Equivalence &amp; Magnetic Equivalence, which is often misinterpreted. In the first example you have given, all the protons don't have the same chemical shifts and this is seen from the multiplet pattern as well. What you can obviously predict is that the two ortho protons are chemically equivalent, and the two meta protons are also chemically equivalent, and the para proton is a separate different type of proton. For observing these differences, what you need to do is to record the NMR at high frequency (e.g., $500$ or $\pu{600 MHz}$). There you can see the three different peaks clearly. Now in this case only, the two ortho protons are also magnetically equivalent as they couple with all the other protons equally, i.e. the $^3\!J_{H-H}$ and $^4\!J_{H-H}$  values are same for the two ortho protons, but in general this may not be the case at all (which is simply to consider another different para substitution at the ring, then the two ortho protons will become magnetically non-equivalent and therefore will have different chemical shift which is observable). Similar is the case with meta potons i.e. they are also magnetically equivalent. So those two protons will have same chemical shift. But ortho, meta, and para protons will definitely have different chemical shifts in your first example. If you zoom in or record the spectra in a high frequency machine, you should see a doublet of doublet peak (due to $^3\!J_{H-H}$ and $^4\!J_{H-H}$) for both ortho and meta protons, and a triplet of triplet for the para proton.

In the second example also, the two protons ortho to $\ce{-N(CH3)2}$ are chemically as well as magnetically equivalent as they both couple with the remaining proton in the same way. So, applying the similar logic you should see a doublet peak (due to $^4\!J_{H-H}$ only) for those two protons and a triplet peak for the remaining one and these two peaks should be at two different chemical shifts. 

To encourage you more here are some spectra from the book Introduction to Spectroscopy which emphasize on aromatic regions and consider different spin-spin splitting and show how complicate a structure can be: 



The above spectra depicts the differences in the spectra between ortho, meta, and para substitution in the benzene ring. The below two are some typical examples of complicated spectra due to spin-spin splitting as well as magnetic non-equivalence. 





From these examples you can see that the aromatic protons can definitely come at distinct regions and what you need is to just record them in high field NMR machine and zoom into that region.