 Solution

For every odd prime $p$ and $n\ge 1$, the group of units modulo an integer $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$ is a cyclic group, of order $\phi (p^n)=p^{n-1}(p-1)$; a generator is a primitive root.

Modulo seven, $3,3^2,\dots ,3^6$ are $3,2,6,4,5,1$, so three has order six. Moreover $3^6-1=728=7\cdot 104$ is divisible by seven exactly once. We prove the lifting needed here. If $u=1+7^{r}v$ with $r\ge 1$ and $7\nmid v$, the binomial expansion gives $u^7=1+7^{r+1}v+O(7^{r+2})$, so $v_7(u^7-1)=r+1$. Induction therefore givesAny exponent giving one modulo $7^n$ is divisible by six. The order also divides $6\cdot 7^{n-1}$ by Lagrange theorem, while the displayed valuation rules out each proper divisor of this form. Consequently three is a primitive root modulo every $7^n$.

Since $2^3=1(\mathrm{mod}7)$ and two is not one, its order modulo seven is three. A generator modulo $7^n$ would reduce to a generator modulo seven, because reduction is a surjective group homomorphism. Thus two is never a primitive root modulo $7^n$. Finally every unit modulo eight is odd, and $(2k+1{)}^2=1+4k(k+1)\equiv 1(\mathrm{mod}8)$. All unit orders are at most two, whereas the unit group has four elements. There is no primitive root modulo eight.