Solution: We recall that if $A$ is a nonepty subset of $\mathbb{R}$ is bounded, then its supremum and infimum exist.

1. Since the set $A$ is non-empty and bounded, the supremum and infimum of $A$ will exists. Since $A$ is bounded, $-A$ will also be bounded and hence $-A$ has also the supremum and the infimum. Let $\sup(-A) = -\alpha $. We need to show that \[ \inf A = \alpha . \] Note that since $-\alpha $ is the supremum of $-A$, so it is an upper bound of $-A$ and therefore, for all $a\in A$, we have \[ -\alpha \geq -a \implies \alpha \leq a , \] which implies $\alpha $ is a lower bound of $A$. Now it remains to show that $\alpha $ is the infimum, that is if $\beta $ is another lower bound of $A$, then $\beta \leq \alpha $. Since $\beta $ is a lower bound of $A$, so for all $a\in A$ \begin{align*} \beta \leq a \implies -\beta \geq -a, \end{align*} which implies $-\beta $ is an upper bound for $-A$. As $-\alpha $ is the supremum of $-A$, so we must have \[ -\alpha \leq -\beta \implies \alpha \geq \beta . \] Therefore, we showed that \[ \sup(-A) = -\inf (A). \]

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2. This is again similar to the previous one. Let $-\alpha =\inf(-A)$. We need to show that $\sup(A) = \alpha $. As $-\alpha $ is an infimum of $-A$, so a lower bound of $-A$, thus for all $a\in A$ we have \[ -\alpha \leq -a \implies \alpha \geq a, \] which implies $\alpha $ is an upper bound of $A$. Now we need to show that it is the least upper bound (supremum) of $A$. If $\beta $ is another upper bound of $A$, then for all $a\in A$ \[ a\leq \beta \implies -a \geq -\beta , \] which implies $-\beta $ is a lower bound of $-A$ and hence \[ -\beta \leq -\alpha \implies \beta \geq \alpha . \] Thus, \[ \inf(-A) = -\sup(A). \]