I have two problems quite similar. The first:
In $\mathbb{Z}_8$ find the identity of the following commutative operation: $\overline{a}\cdot\overline{c}=\overline{a}+\overline{c}+2\overline{a}\overline{c}$
I say: $\overline{a}\cdot\overline{i}=\overline{a}+\overline{i}+2\overline{a}\overline{i} = \overline{a}$ Since $\overline{a}$ is always cancelable in $\mathbb{Z}_9$ I can write: $\overline{i}+2\overline{a}\overline{i} = \overline{0}$ $\overline{i}(\overline{1}+2\overline{a}) = \overline{0}$ so $i=0$ whatever $\overline{a}$.
Second question:
In $\mathbb{Z}_9\times\mathbb{Z}_9$ find the identity of the following commutative operation: $(\overline{a}, \overline{b})\cdot (\overline{c}, \overline{d})= (\overline{a}+ \overline{c}, \overline{8}\overline{b}\overline{d})$
So starting from: $(\overline{a}, \overline{b})\cdot (\overline{e_1}, \overline{e_2})= (\overline{a}+ \overline{e_1}, \overline{8}\overline{b}\overline{e_2})=(\overline{a}, \overline{b})$ that is: $\overline{a}+\overline{e_1}=\overline{a}\qquad (1)$ $\overline{8}\overline{b}\overline{e_2}=\overline{b}\qquad (2)$ In (1) there's always cancellable element for $\overline{a}$ since $\overline{-a}$ is always present in $\mathbb{Z}_9$. In (2) I should multiply both member for $\overline{8^{-1}}$ and $\overline{b^{-1}}$ to know exactly $\overline{e_2}$. This happens only if both number are invertible. $\overline{8}$ is easily to demonstrate that it's invertible, cause $gcd(8,9)=1$. But what about $b$.