Is [Co(NH₃)₄Cl₂]Cl paramagnetic or diamagnetic?
The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $\ce{Co^3+}$ is $\mathrm{d^6}$, and virtually all "octahedral" $\mathrm{d^6}$ complexes are low spin - essentially some complexes of $\ce{Fe^2+}$  and a very small number of fluoro complexes of $\ce{Co^3+}$ are the only exceptions to the rule that all $\mathrm{d^6}$ octahedral complexes are low spin.

Why is this? Well let's use a bit of Crystal Field Theory (much of below stolen from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy). One can consider the crystal field stabilization energy which is how stabilized an ion is compared to the free ion when in a given electronic state electronic and geometry. Here we are interested in high and low spin, and octahedral geometry. Taking a table of these values from the above link:



In the above $\Delta_\circ$ is the splitting energy of the orbitals. We can see that low spin $\mathrm{d^6}$ has the largest possible stabilization energy of any electronic configuration compared to the high spin case ($2\Delta_\circ$-P). Thus unless the splitting is very small octahedral $\mathrm{d^6}$ prefers low spin. So how do we get small splitting to see high spin $\mathrm{d^6}$? Well small splitting is favoured by


Low charge first row transition metals - for instance $\ce{Fe^{2+}}$
Very weak field ligands 


And what were our exceptions above?