Let $A$ be a $3 \times 3$ real matrix satisfying $A^8 = I$. Which of the following is true?
Minimal polynomial of $A$ is of degree $3$
Minimal polynomial of $A$ is of degree $2$
$A = I$ or $A = -I$
There are uncountably many such $A$
My thoughts: the minimal polynomial divides $(x^4-1)(x^4+1)$ and is itself of degree $\leq 3$, so it can be of degree $2$ or $3$.
This is not technically an answer, but more of a longer comment
Like you said, the minimal polynomial of $A$ divides $x^8 - 1$. Observe that $$x^8-1 = (x-1)(x+1)(x^2+1)(x^4+1). $$
Moreover, you can further decompose $$x^4 + 1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1), $$ and the polynomials on the RHS are irreducible in $\mathbb{R}$.
Thus, if $A$ has only rational elements, then the minimal polynomial of $A$ divides $(x-1)(x+1)(x^2+1)$. However, the minimal polynomial of $A$ cannot be $x^2+1$, because, if $A^2 = -I$, then $A$ would be an $n \times n$ square matrix, with $n$ an even number (look at the Jordan normal form $J$ of $A$; you would have that $J^2 = -I$). So the minimal polynomial is either $x-1, x+1, (x-1)(x+1), (x-1)(x^2+1)$ or $(x+1)(x^2+1)$, and I believe all of them are possible.
However, if $A$ has real elements, the minimal polynomial can also contain factors of $x^4+1$ (but not $x^4+1$ itself, like you observed).
However, what I said technically implies that 1), 2) and 3) are false and that 4) is true (you can also see the comments).