Basing myself from a similar question, I was able to get this:
Since $\phi(8)=4$ then:
$3^4 \equiv 81(\mod41) \rightarrow 3^4 \equiv -1 (\mod41) $ then $3^8 \equiv 1 (\mod 41)$
Now this is the part I don't understand.
$\{3,27,38,14\}$ are primitive $8^{th}$ roots of unity $\mathbb{Z}/41\mathbb{Z}$
I know how they got those values since:
$3^3 \equiv 27 (\mod 41)$
$3^5 \equiv 38 (\mod 41)$
$3^7 \equiv 14 (\mod 41)$
I was able to get the first primitive $8^{th}$ root, but I don't understand why the other ones are also primitive roots. As you can see I have the answers, but I want to understand how they got them.
Edit: So I understood the process of why those were the answers. The only one i'm getting stuck is on $3^2$ since $ 3^2 \equiv -9(\mod 41)$ which is not $\equiv 1 (\mod 41)$ Thus it should also be a primitive root? For example $3^3 \equiv 27 (\mod 41)$. Then $27^1 \not\equiv 1$, $27^2 \equiv 32 (\mod 41)$ Thus $32 \not\equiv 1$ So 27 is a primitive root. So why wouldn't 9 be considered a primitve root? If possible can someone give me an example to work on?
Being $g$ a generator of $\mathbb F^*_{41}$ (i.e. a primitive root modulo $41$) we have $$\alpha=g^{\frac{41-1}{8}}=g^5\Rightarrow \alpha^8=g^{40}=1 \space\space\text {(by FLT) }$$ and $$\alpha^k=g^{5k}\ne1\space\space\text {for }0\lt k\lt8\space\space\text {because g is a generator}$$ Consequently we have to calculate $g^5$ for all generator of $\mathbb F^*_{41}$
There are $16$ primitive roots modulo $41$ which are
$6,7,11,12,13,15,17,19,22,24,26,28,29,30,34,35$ Calculation gives $$6^5=17^5=19^5=26^5=27\\7^5=13^5=29^5=30^5=38\\11^5=12^5=28^5=34^5=3\\15^5=22^5=24^5=35^5=14$$ Thus there are four solutions $$\color{red}{3,14,27,38}$$