I suppose your definition of Dedekind cut is: $A$ is a proper subset of $\mathbb Q$ which is downwards closed and does not have largest element.
It seems that you assume $\alpha>0$ (i.e., $\{p\in\mathbb Q; p\le 0\}\subseteq\alpha$), even though you did not mention this in your question.
Then $\alpha^{-1}$:
- is a proper subset of $\mathbb Q$, since $p\mapsto\frac1p$ is a bijection $\mathbb Q^+ \to \mathbb Q^+$
- is downwards closed: Suppose $p\in\alpha^{-1}$ and $q
. If $p\le 0$, then $q\le 0$ and $q\in\alpha^{-1}$. If $p>0$ then we have $\frac1p\in\alpha^c$, and the inequality $\frac1q>\frac1p$ implies $\frac1q\in\alpha^c$ (since $\alpha^c$ is upwards closed) and it is not the smallest element of $\alpha^c$.
- does not have largest element. Suppose, by contradict, that $p$ is the largest element of $\alpha^{-1}$. This means $p>0$ and since $p\mapsto\frac1p$ is an order-reversing map on $\mathbb Q^+$, this would imply that $\frac1p$ is the smallest element of $\alpha^c$.
For question 2: I think it should be relatively easy to show that if $\alpha=\{p\in\mathbb Q; p for some $r\in\mathbb Q^+$, then $\alpha^{-1}=\{p\in\mathbb Q; p<\frac 1r\}$.
(By $\mathbb Q^+$ I mean the set $\{q\in\mathbb Q; q>0\}$.)