Solutions

a) Let $U$ be the set of almost upper bounds and $L$ the set of almost lower bounds.

i) $U=(0,\infty)$, $L=(-\infty,0]$.
ii) $U=(0,\infty)$, $L=(-\infty,0)$.
iii) $U=(0,\infty)$, $L=(-\infty,0]$.
iv) $U=[\sqrt{2},\infty)$, $L=(-\infty,0]$.
v) $U=\varnothing$, $L=\varnothing$.
vi) $U=\left[\frac{-1+\sqrt{5}}{2},\infty\right)$, $L=\left(-\infty,\frac{-1-\sqrt{5}}{2}\right]$.
vii) $U=[0,\infty)$, $L=\left(-\infty,\frac{-1-\sqrt{5}}{2}\right]$.
viii) $U=(1,\infty)$, $L=(-\infty,-1]$.

b) Since $A$ is bounded, it is also bounded above. This means that it has an upper bound. This upper bound is in $B$, so $B$ is nonempty.

For contradiction, suppose $B$ were not bounded below. Let $l$ be a lower bound of $A$. Since $B$ is not bounded below, there exists $b\in B$ such that $b<l$. But by the definition of $B$, this would imply that there are only finitely many elements of $A$ greater or equal to $b$. This contradicts the premise that $A$ is infinite.

c)

i) $\overline{\lim}A=0$.
ii) $\overline{\lim}A=0$.
iii) $\overline{\lim}A=0$.
iv) $\overline{\lim}A=\sqrt{2}$.
v) Does not exist.
vi) $\overline{\lim}A=\frac{-1+\sqrt{5}}{2}$.
vii) $\overline{\lim}A=0$.
viii) $\overline{\lim}A=1$.

d) We define $\underline{\lim}A$ as the lowest upper bound of the set of all almost lower bounds of $A$.

i) $\underline{\lim}A<0$.
ii) $\underline{\lim}A=0$.
iii) $\underline{\lim}A=0$.
iv) $\underline{\lim}A=0$.
v) Does not exist.
vi) $\underline{\lim}A=\frac{-1-\sqrt{5}}{2}$.
vii) $\underline{\lim}A=\frac{-1-\sqrt{5}}{2}$.
viii) $\underline{\lim}A=-1$.