How to construct isomorphisms between finite field? Well, Problem says: 

The polynomial $p_{1}(x)=x^{2}+x+1\in\mathbb{F}_{5}[x]$ is
  irreducible. Call $\theta$ an element in
  $\mathbb{F}_{25}=\mathbb{F}_{5}[x]/(p_{1}(x))$ that satisfies
  $\theta^{2}+\theta+1=0.$
(a) Find all other irreducible monic quadratic polynomials in
  $\mathbb{F}_{5}[x]$. 
(b) For each of the $10$ polynomials found in the previous part, write
  the two roots in $\mathbb{F}_{25}$ as $a\theta+b$ for suitable
  $a,b\in\mathbb{F}_{5}.$

First part is not so hard since there are only 25 elements in $\mathbb F_5[x]$. But the second part seems to me to hard since I think I have to construct explicitly isomorphisms between $\mathbb F_{25}$ and $\mathbb F_5[x]/(f(x))$ where $f(x)$ is an irreducible monic quadratic polynomials in $\mathbb F_5[x]$. 
For example, take $f_1(x)=x^2+2$. Clearly it is an irreducible monic quadratic polynomials in $\mathbb F_5[x]$. Then I have to construct isomorphism between $\mathbb F_{25}$ and $\mathbb F_5[x]/(x^2+2)$. How can I do it?
 A: I suggest approaching the problem in reverse, making substitutions of the form $x=ay+b$ in the polynomial $x^2+x+1$.  Choose explicit values from $\mathbb F_5$ for $a$ and $b$.  Never choose zero for $a$.  The polynomials in $y$ you obtain will be irreducible, since if $\theta$ generates $\mathbb F_{25}$ then so does $a^{-1}(\theta - b)$.
As an example, consider the substitution $x=y+2$.  The polynomial $x^2+x+1$ becomes $y^2+2$.  One  root of $y^2+2$ is $(\theta-2)$. The other is $(-\theta -3)$ since  $\text{Gal}(\mathbb F_{25}/\mathbb F_5)$ is generated by the map that sends each element to its fifth power, and since $\theta^5 = \theta^2 = -\theta - 1$.
A: Another method relies, similarly to what is done in this other answer, on reversing the order of the questions.
First note that the Frobenius automorphism $u \mapsto u^ 5$ maps $\theta$ to $\theta^{5} = - \theta - 1$.
Now consider all the $20$ elements of $\mathbb{F}_{25} \setminus \mathbb{F}_{5}$. These are the $u = a \theta + b$, with $a, b \in  \mathbb{F}_{5}$, $a \ne 0$.
Your irreducible polynomials will be the 
$$
(x - u) (x - u^5)
=
(x - (a \theta + b)) (x - (-a \theta + b - a))
=
x^{2} - (2 b - a) x + b^{2} - a b + a^{2}.
$$
You can get each of the $10$ irreducible polynomials uniquely by choosing $b \in \mathbb{F}_{5}$ and $a = 1, 2$.
A: Any element of the form $r=a+b\theta$, with $b\ne0$, has degree $2$ over $\mathbb{F}_5$. We have $r-a=b\theta$, so
$$
(r-a)^2=b^2\theta^2=-b^2\theta-b^2=-b(r-a)-b^2
$$
so the minimum polynomial for $r$ is
$$
X^2+(b-2a)X+a^2-ab+b^2
$$
Substituting all possible values gives the irreducible degree $2$ polynomials (with repetitions).
We can use the fact that, if $a+b\theta$ is a root of a polynomial with coefficients in $\mathbb{F}_5$, then also its conjugate $a+b(-1-\theta)=(a-b)-b\theta$ is.
For instance, $a=0$ and $b=1$ gives $a-b=4$ and $-b=4$, so we can avoid computing the polynomial corresponding to $a=4$ and $b=4$:
\begin{array}{c|c|l}
(a,b) & (a-b,-b) & \text{Polynomial} \\
\hline
(0,1) & (4,4) & X^2+X+1 \\
(1,1) & (0,4) & X^2+4X+1 \\
...
\end{array}
