I have read over the sections in my book and came to following conclusions, that I hope are correct...
Starting with the compound with the two bromines and going clockwise...
If we are using the compound in the middle to make the compound at the top, I think the double bond would attack Br2 (my theory for the reagent). This would split the Br2 into two radicals. Since the double bond has broken, there are two radicals on the compound as well; allowing room for the bromines to attach onto the compound.
The next compound with no substituents hanging off, I believe the reagent to be H2 for the same reason as above.
The next two compounds I believe use HBr because I believe that the double bond breaks the HBr, and one H goes to one radical that came from the broken bond while the Br goes to the other radical on the compound.
The compound with the OH and Br, I believe the first reagent is HBr and then OH because the double bond will break the HBr, allowing the Br to attach to the compound, and then the OH instead of the H will attach to the other part of the compound where the other radical is located.
The compound with the OCH3, I believe the reagent is OCH3
The last two compounds with the OH only, I believe has a reagent of OH for the same reason as the two compounds with the Br.
Any imput on my theories would be greatly appreciated!