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 $
