2's complement of $F2_{(16)}$ in 8 bits

Question

I found an exercise asking the representation of $F2_{(16)}$ in 2's complement with 8bits.

I've made de conversion to binary: $11110010_{(2)}$

And I followed the procedure:

1. transform the $1$'s to $0$'s and $0$'s to $1$'s;
2. add $1$;

and got: $00001110$

But as I was doing it I was thinking $F2_{(16)}$ is $242$ and using 2's complement and 8 bit let's me represent numbers up to $127$. Then I went to several online converters and some showed the result I got and others that it was impossible since the convertion would need 9bits.

My question is: is it possible that the question is wrong or am I just not correct about the number limit that can be represented in this way? (I believe there are only 255 different numbers that can be represented)

EDIT: Taking into account this is a translation, the exercise is as follows: Which are the integers represented in 2's complement with 8 bits by $F2_{16}$, $157_{8}$, $4F_{16}$, $10010000$, $00000000$? Is it possible that I'm not even interpreting the exercise correctly?