$x^2\equiv 1\ (\mod 8)\\
x^2-1\equiv0\ (\mod 8)\\
(x-1)(x+1)\equiv0\ (\mod 8)\\
8\mid(x-1)(x+1)
$
By trial and error, we get $x\equiv a\ (\mod8)$, where $a\in\{1,7,3,5\}$.
In modular arithmetic, all elements are members of an equivalence class and as such, referencing any member of a class is an acceptable way to represent that class.
For example, modulo $45$, the element $28$ is a representative of the equivalence class comprising integers of the form $28 + 45k$, where $k$ is any integer, where the equivalence relation $\equiv$ between two elements $a$ and $b$ is $45 \mid (a-b)$; i.e., their difference is divisible by $45$.
Because of this, there is no such thing as a unique representative of the set of equivalence classes; you can choose any representation you wish. The representation $\{0, 1, 2, \ldots, 44\}$ is in some sense "canonical," but it is not forced, because in referring to a class by a representative element, say $37$, we are not speaking of the specific element itself, but the entire class it represents.
Consequently, to accept $37$ as an answer but not $-8$ is mathematically incorrect. Both are multiplicative inverses of $28$ modulo $45$ since as was previously noted, $(37)(28) = 1036$ and $(-8)(28) = -224$, both of which are members of the same equivalence class $1$, modulo $45$. In order to make such a preference acceptable, the desired group representation must be explicitly stated in advance; e.g., "find the least nonnegative integer $a$ that $28a \equiv 1 \pmod {45}$," or something to that effect. Otherwise, simply saying "find the multiplicative inverse of $28$ modulo $45$" is misleading because the word "the" erroneously implies such an inverse is unique when in fact any member of the corresponding equivalence class is a solution.
Best Answer
The easiest way is to notice that $5x = 0$ or $5 \mod 10$. So it can never be 8. This can be shown formally.
If $x$ is even then let $x=2k$, then $5x = 10k \equiv 0 \neq 8 \mod 10$.
If $x$ is odd then let $x=2k+1$, then $5x = 10k+5 \equiv 5 \neq 8 \mod 10$.
Therefore the equation $5x \equiv 8 \mod 10$ has no solutions.