Does coNP-completeness imply NP-hardness?
You claim that every problem in NP can be reduced to its complement, and this is true for Turing reductions, but (probably) not for many-one reductions. A many-one reduction from $L_1$ to $L_2$ is a polytime function $f$ such that for all $x$, $x \in L_1$ iff $f(x) \in L_2$.

If some problem $L$ in coNP were NP-hard, then for any language $M \in NP$ there would be a polytime function $f$ such that for all $x$, $x \in M$ iff $f(x) \in L$. Since $L$ is in coNP, this gives a coNP algorithm for $M$, showing that NP$\subseteq$coNP, and so NP$=$coNP. Most researchers don't expect this to be the case, and so problems in coNP are probably not NP-hard.

The reason we use Karp reductions rather than Turing reductions is so that we can distinguish between NP-hard and coNP-hard problems. See this answer for more details (Turing reductions are called Cook reductions in that answer).

Finally, coNP-hard and coNP-complete are both standard terminology, and you are free to use them.