The second problem: The coefficients have only 3 possibilities, 0,1 or 2. And the power of $x$ in the quotient ring by the ideal of $x^3-x-1$ has to be at most two and so the quotient ring will have $3\times3\times3=27$ elements. Considering you want to avoid any concept like irreducibility you can simply write the multiplication table for all the non-zero elements and realise 1 as a product making everything invertible, hence a field.
For example constants have inverses, and in the quotient ring $x^3-x-1=0$ which is the same as saying $x^3-x=1$ or $x$ and $x^2-1$ are inverses of each other.
And $(x^2-1)^2$ will be the inverse of $x^2$.
First problem is also easier: you have just 4 elements in the quotient ring and need to carry out all multiplications and see that product of non-zero elements yield again a non-zero element.
After doing all these and convincing yourself that those ideals are prime and maximal you will realize that alternatives such as Euclidean domain concepts may be worth learning in order to avoid massive calculations.