(a) $C$ is symmetric iff for all $a,b$ , $aCb\Leftrightarrow bCa$ iff for all $a,b$ , $bDa\Leftrightarrow aDb$ iff $D$ is symmetric.

(b) If $C$ satisfies comparability and nonreflexivity, so does $D$ . If $C$ is transitive, then for all $a,b,c$ , $aDb$ and $bDc$ imply $cCb$ and $bCa$ , which imply $cCa$ , which implies $aDc$ , so $D$ is also transitive.

(c) Denote $C$ as $<$ and $D$ as $>$ . Then, using (b) to argue that $(A,<)$ is an ordered set iff $(A,>)$ is an ordered set, if $(A,<)$ has the greatest lower bound property, then for every $B\subset A$ such that $\exists x:x\le b$ for all $b\in B$ , there is $a\in A$ such that $a\le b$ for all $b\in B$ , and for every $d$ such that $d\le b$ for all $b\in B$ , $d\le a$ , then for every $B\subset A$ such that $\exists x:b\ge x$ for all $b\in B$ , there is $a\in A$ such that $b\ge a$ for all $b\in B$ , and for every $d$ such that $b\ge d$ for all $b\in B$ , $a\ge d$ , then $(A,>)$ has the least upper bound property, then (using Exercise 13) $(A,>)$ has the greatest lower bound property, then (using the same argument as above for $(A,>)$ ) $(A,<)$ has the least upper bound property.

I don’t know if this looks any easier than a direct argument similar to Exercise 13 would do.