Finding a maximal ideal and a prime ideal in $\mathbb Z_8[x]$ 
$1.$ Find a maximal ideal and a prime ideal in $\mathbb Z_8[x]$

Attempt:  Finding a maximal ideal, I am not sure how do I go about it. $\mathbb Z_8[x]$ is not a $PID$, so there's no use finding an irreducible polynomial as well.
EDIT:   I am trying to show $2$ must be in all prime/maximal ideals
What should be the logical procedure in actually trying to find out a maximal ideal in any ring?

$2.$ Let $R=\{a/b~~|~~a,b \in \mathbb Z, 3 \nmid b\}$ . We need to find it's field of quotients.

Attempt: $R$ can be verified to be an integral domain, so, there exists a field $F$ called the field of quotients of $R$ which has a subring isomorphic to $D$
$R$ is not a field as $3/2$ is not invertible in $R$.
Let $p,q \in R$. Then let $p/q$ denote the equivalence class containing $(p,q)$
Then define $p/q + r/s = ps+rq/qs$ and $p/q \cdot r/s =pr/qs$
The set of these quotients form a field isomorphic to the set of rational whose denominators are not a multiple of $3$.
Am I correct?
Thank you for your help.
 A: Hint for #1. Show that $2$ must be in all prime ideals. Also convince yourself of the fact that $\Bbb{Z}_8[x]/\langle 2\rangle\cong\Bbb{Z}_2[x]$. Can you find prime ideals/maximal ideals in the latter ring, and pull them back to the original ring?
Hint for #2. Can you think of a field that contains $R$? What is the smallest such field? Can you show that it is the field of quotients of $R$.
A: Lots of questions here, so I'll focus on just one of them.  One way of finding a maximal ideal in $\mathbb{Z}_8[x]$ is using the fact that an ideal $I$ is maximal if and only if $\mathbb{Z}_8[x]/I$ is a field.
First, recognize that $\mathbb{Z}_8/\langle 2 \rangle \cong \mathbb{Z}_2$, and so it follows that $\mathbb{Z}_8[x]/\langle 2 \rangle \cong \mathbb{Z}_2[x]$.  What are the maximal ideals in $\mathbb{Z}_2[x]$?  The corresponding maximal ideal in $\mathbb{Z}_8[x]$?  Hint: The latter won't be principal.
I should add that all maximal ideals are prime ideals, so you've really answered two questions if you can answer this one. 
Edit: If you wish to find an ideal that is prime, but not maximal, then apply the fact that an ideal $I$ in a ring $R$ is prime $\iff$ $R/I$ is an integral domain.  Another hint for approaching this: $R/I$ cannot be finite since all finite integral domains are fields.
