Factor x^{4}-2 into irreducible polynomials.

Step 1
You had a wrong sign in the factorization in ${\displaystyle \mathbb{R}\left[x\right]}$ (typo):
$(x^2+\sqrt{2})(x-\sqrt[4]{2})(x-\sqrt[4]{2})$
is the factorization: the quadratic factor has no roots in ${\displaystyle \mathbb{R}}$, so it's irreducible.
The polynomial is indeed irreducible in ${\displaystyle \mathbb{Q}\left[x\right]}$: it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in ${\displaystyle \mathbb{R}\left[x\right]}$ and the only way you can combine the previous factorization in the required way would be ${\displaystyle (x^2+\sqrt{2})(x^2-\sqrt{2})}$ and the factors are not in ${\displaystyle \mathbb{Q}\left[x\right]}$. Or you can use Eisenstein's criterion, which applies to ${\displaystyle x^4-2}$ for the ' 2.
Sorry, but the factorization in ${\displaystyle \mathbb{C}\left[x\right]}$ is wrong: it should be a product of linear factors and you get them from the factorization in ${\displaystyle \mathbb{R}\left[x\right]}$
I'll do the case ${\displaystyle {\mathbb{Z}}_3\left[x\right]}$. The polynomial has no roots in ${\displaystyle {\mathbb{Z}}_3}$, but it could be a product of two (monic) quadratic factors:
${\displaystyle x^4-2=(x^2+ax+b)(x^2+cx+d)}$
Since ${\displaystyle bd=-2=1}$, you can have ${\displaystyle b=1}$ and ${\displaystyle d=1}$ or ${\displaystyle b=2}$ and ${\displaystyle d=2}$. In the first case we have.
${\displaystyle (x^2+ax+1)(x^2+cx+1)=x^4+(a+c)x^3}$
${\displaystyle +(2+ac)X^2+(a+c)x+1}$
and so you need ${\displaystyle a+c=0}$ and ${\displaystyle ac+2=0}$. Hence ${\displaystyle c=-a}$ and ${\displaystyle a^2=2}$, which is impossible.
In the second case we have
${\displaystyle (x^2+ax+2)(x^2+cx+2)=x^4+(a+c)x^3+(1+ac)x^2+(2a+2c)x+1}$
Again ${\displaystyle a+c=0}$ and ${\displaystyle 1+ac=0}$. Since ${\displaystyle c=-a}$, we obtain ${\displaystyle a^2=1}$. Well, this is indeed possible: we have ${\displaystyle a=1}$ or ${\displaystyle a=2}$. Thus the factorization is
${\displaystyle (x^2+x+2)(x^2+2x+2)}$

Step 1
The ${\displaystyle {\mathbb{Z}}_{17}}$ case remains. The units of ${\displaystyle {\mathbb{Z}}_{17}}$ form a cyclic group of order 16. The order of 2 is 8, as ${\displaystyle 2^4=-1}$. Thus ${\displaystyle \sqrt{2}\in {\mathbb{Z}}_{17}}$, and your polynomial factorises as:
${\displaystyle x^4-2=(x^2+\sqrt{2})(x^2-\sqrt{2})}$
It cannot factorise further as if there were a linear factor, then ${\displaystyle {\mathbb{Z}}_{17}}$ would contain an element ${\displaystyle \alpha }$ with ${\displaystyle {\alpha }^4=2}$, so ${\displaystyle {\alpha }^{16}=-1}$, which is impossible as for any element ${\displaystyle \alpha \in {\mathbb{Z}}_{17}}$ we have ${\displaystyle {\alpha }_{16}=1}$.
To complete the question you should work out ${\displaystyle \sqrt{2}}$ in ${\displaystyle {\mathbb{Z}}_{17}}$. The easiest way to do this is check values of ${\displaystyle 2+17k}$, for ${\displaystyle k=0,1,2,3,\cdots }$ till you get a perfect square.