Some questions on the ring $(\mathbb{F}_3,+,\cdot)$ 
*

*Find an irreducible polynomial in $\mathbb{F}_3$ of degree 2.


The polynomials to analyse are $9$ in the form:
$X^2+aX+b$ where $a,b \in \mathbb{F}_3$
I find out that the irreducible ones are:
$$X^2+1$$
$$X^2+X+2$$
$$X^2+2X+2$$


*For the polynomial $p(X)$ found above find the multiplicative inverse to $X+1+p(X)\mathbb{F}_3$ in the quotient ring $(\mathbb{F}_3[X] / p(X)\mathbb{F}_3[X],+,\cdot)$


I pick $p(X)=X^2+1$
$(X+1 + \langle X^2+1\rangle)(aX^2 + bX + c + \langle X^2+1\rangle) = 1 + \langle X^2+1\rangle  \Rightarrow \\  
\Rightarrow aX^3 + bX^2 + cX + aX^2 + bX + c + \langle X^2+1\rangle = 1 + \langle X^2+1\rangle \Rightarrow  \\
\Rightarrow aX^3 + bX^2 + cX + aX^2 + bX + c + \langle X^2+1\rangle = 1 + \langle aX^3+X^2+1\rangle   \Rightarrow  \\
\Rightarrow aX^3+ (a+b)X^2 + (b+c)X + c + \langle X^2+1\rangle = 1 + \langle X^2+1\rangle$
I know that $c=1$ and:
$$a=0\mod3$$
$$a+b=0\mod3$$
$$b+c=0\mod3$$
but I guess there is a mistake :/


*Is $X+1+p(X)\mathbb{F}_3$ a primitive element of $R=(\mathbb{F}_3[X] / p(X)\mathbb{F}_3[X],+,\cdot)$?


For this point I thought that a primitive element of $R$ should have order $3^2-1=8$ 
Any element of $R$ has an order dividing $8$, that could be $1,2,4,8$.
I just have to check the order of $X+1+p(X)\mathbb{F}_3$, and if it is $8$ therefore it is a primitive element of $R$ 

Your process for finding the multiplicative inverse is wrong. In particular, in the presence of the ideal $\langle X^2+1\rangle$, your implication that $a=0$ (for instance) is completely wrong - for instance, $X^3+X^2+1 + \langle X^2+1\rangle = 1+\langle X^2+1\rangle$.
You're better off first showing that every polynomial in $X$ is equivalent $\pmod{\langle X^2+1\rangle}$ to a polynomial of the form $aX+b$ and then doing the multiplication on that polynomial (using the ideal to reduce appropriately).
