Show $2+\alpha$ is a primitive root of $\mathbb{F}_{25}$. Suppose $\alpha \in \mathbb{F}_{25}$ is an element with $\alpha^2 = 2$, I need to prove that $2+\alpha  \in \mathbb{F}_{25}$ is a primitive root (that is: a generator of the cyclic group $\mathbb{F}_{25}^\ast$ of order 24).
So far, I have tried the following. 
Since $\mathbb{F}_{25} = \mathbb{F}_{5^2}$, consider $f = X^2+X+2 \in \mathbb{F}_5[X]$. This is irreducible since it has no roots in $\mathbb{F}_5$. Furthermore we have
$$
f(2+\alpha) = (2+\alpha)^2+(2+\alpha)+2 = \alpha^2 + 4\alpha + 4 + 2 + \alpha + 2 
= \alpha^2 + 3 = 5 = 0. 
$$
So it follows that $f$ is the minimum polynomial of $2+\alpha \in \mathbb{F}_{25}$, hence
$$
\mathbb{F}_{5}(2+\alpha) \cong \mathbb{F}_5[X]/(X^2+X+2) \cong \mathbb{F}_{25}. 
$$
How to proceed from here though? 
 A: Suppose that $2+\alpha$ is not a primitive root; then its multiplicative order is a proper divisor of $24$, so it divides either $8$ or $12$. Compute $(2+\alpha)^8$ and $(2+\alpha)^{12}$. It helps to first compute $(2+\alpha)^4$.

On a side note; your efforts to show that $\Bbb{F}_5(2+\alpha)\cong\Bbb{F}_{25}$ seem wasted to me. After all it is clear that $\Bbb{F}_5(2+\alpha)=\Bbb{F}_5(\alpha)$, and that $\Bbb{F}_5(\alpha)=\Bbb{F}_{25}$ because $2$ is not a square in $\Bbb{F}_5$. At best, your finding that $2+\alpha$ is a root of $X^2+X+2$ helps to show that $2+\alpha$ is a root of the $24$th cyclotomic polynomial $\Phi_{24}$, which factors over $\Bbb{F}_5$ as
$$\Phi_{24}(X)=X^8-X^4+1=(X^2+X+2)(X^2+X+3)(X^2+3X+3)(X^2+4X+2).$$
I don't know of an easy way to find that factorization by hand, and plugging in $2+\alpha$ into $\Phi_{24}(X)=X^8-X^4+1$ is about as much work as computing $(2+\alpha)^8$ and $(2+\alpha)^{12}$.
A: I can't resist pointing out the following alternative that in a way allows us to use Moivre's formula here.
Observe that in $\Bbb{F}_5$ we have $2=-3$. Therefore
$$
z=2+\alpha=-3+\sqrt{-3}=-3+ i\sqrt3
$$
for a suitable meaning of $i$. When viewed as a complex number $z$ has absolute value $2\sqrt{3}$ and argument $5\pi/6$. Therefore $z^3$ has absolute value $24\sqrt3$ and argument $\pi/2$. In other words (for the same choice of $i$ as above)
$$
z^3=24i\sqrt3 \equiv-i\sqrt3\implies z^6=-3.
$$
Proceeding from here is easy: $z^{12}=9\equiv-1$ as expected, and
$z^8=-3z^2$ can be reduced using the minimal polynomial you worked out.
