$A :=(a,b) =\{x \in \mathbb{R} \mid a<x<b\}$.
Then lub of $A=b$ and glb of $A = a$.
Clearly $b$ is upper bound for $A$.
Since $\frac{b+a}{2}\in A$.
where $\frac{b+a}{2}$ is midpoint of a and b.
Let $\alpha$ be an upper bound of A.
such that $a<\frac{a+b}{2}< \alpha$.
If $\alpha \geq b $. If not $\alpha<b\implies \alpha \in (a,b)$.
Now if x is midpoint of $\alpha$ and b then $x=\frac{\alpha+b}{2}$.
$x-\alpha =\frac{b-\alpha}{2}>0 \implies b-\alpha>0 \implies b > \alpha$.
That is $\alpha$ is not an upper bound of A.
And Second Question is How to "show that $\operatorname{glb}(a,b)=a$" for the seocnd question if we take any $\beta$ such that $a<\beta<\frac{a+b} {2} < b$
and let $y$ be midpoint of $a$ and $\beta $ then $y=\frac{a+\beta}{2}$. now $\beta -a =\frac{a+\beta}{2}-a=\frac{\beta-a}{2}>0\implies \beta-a>0\implies \beta\geq a $
I reorganized your proofs below.
Let $x=\mbox{l.u.b.}(a,b)$. Since $b$ is an upper bound, we have that $x\leq b$. Assume that $x<b$, then we have two cases.
i) If $x\leq a$ then $x<\frac{a+b}{2}\in(a,b)$ and $x$ is not an upper bound. Contradiction!
ii) If $a<x$ then $x<\frac{x+b}{2}\in(a,b)$ and $x$ is not an upper bound. Contradiction!
Therefore $x=b$.
As regards the $\mbox{g.l.b.}$, the proof is similar.
Let $y=\mbox{g.l.b.}(a,b)$. Since $a$ is a lower bound, we have that $y\geq a$. Assume that $y>a$, then we have two cases.
i) If $y\geq b$ then $y>\frac{a+b}{2}\in(a,b)$ and $y$ is not a lower bound. Contradiction!
ii) If $b>y$ then $y>\frac{a+y}{2}\in(a,b)$ and $y$ is not a lower bound. Contradiction!
Therefore $y=a$.