I did the first one by contradiction and a lot of cases.
Given $ A, B \neq \varnothing \subseteq \mathbf{R}, A+B = \{ a+b \vert a \in A, b \ in B \} $, and I'll let $ \alpha = \sup A, \beta = \sup B $
WTS: $ \sup(A+B) = \alpha + \beta $
First of all, by the lub property of $ \mathbf{R} $ we know that these exist.
Now we show that $ \alpha + \beta $ is an upper bound of $ A+B $ by contradiction. Thus $ \exist a,b \in A,B \ni a+b > \alpha + \beta $
$ (\alpha - a) + (\beta - b) < 0 $
WLOG assume $ (\alpha - a) < 0 $, then $ \exists a \in A \ni a > \alpha, but \alpha = \sup A $, contradiction.
$ \therefore \alpha + \beta $ is an upper bound of $ A+B $
