Let $$E = \frac{\mathbb{Q}[x]}{(x^2+x+1)}$$
For which $p \in \mathbb{N}$ does exist $h \in E$ s.t. $h^2 =p$ or $h^2=-p$
Factorize $x^4-9$ into irreducible polynomials in $E[t]$, where $t$ is an unknow
now, $E$ is a field because $f:= x^2+x+1$ is a cyclotomic polynomial, hence irreducible on $\mathbb{Q}[x]$ and a generic element in $E$ is $h=a+bx + (f)$, where $a,b \in \mathbb{Q}$.
About the first question I tried this path: $h^2 = (a+bx+(f))^2 = a^2+2abx+b^2x^2 + (f) = (2ab-b^2)x+a^2-b^2 + (f)$
placing $\,\,b(2a-b)=0\,\,$ and $\,\,a^2-b^2=p$, but I couldn't get to the conclusion, any hint?
About the second question, I know that $$E[t] = \Bigl\{(a_0+b_0x+(f))+(a_1+b_1x+(f))t+...+(a_n+b_nx+(f))t^n \,\,|\;\;a_i+b_ix+(f)\in E, i\in\{0,...,n\}\Bigl\}$$
but I have no idea how to factorize $x^4-9$ there, any hint?
To continue on your idea for question 1: $b(2a-b)=0$ implies $b=0$, in which case $p=a^2$ is a perfect square. Otherwise $b=2a$ in which case $p=-3a^2$. Because $p$ is integer, $a$ has again to be integer.
For question 2 you can start factorizing in $\mathbb Q[x]$: $$x^4-9=(x^2-3)(x^2+3)$$
Now we know from question 1 that $3$ does not have a square root in $E$, and that $-3$ has. So there exists $\alpha\in E$ s.t. $x^2+3=(x-\alpha)(x+\alpha)$, while $x^2-3$ has to be irreducible. Thus the decomposition is:
$$x^4-9=(x^2-3)(x-\alpha)(x+\alpha)$$
If you let $\beta=x+(f)$ you get from the detail of question 1 that $\alpha=1+2\beta$.