= Solution
For every odd <prime> $p$ and $n\geq1$, the <group of units modulo an integer> $(\mathbb Z/p^n\mathbb Z)^\times$ is a <cyclic group>, of order $\varphi(p^n)=p^{n-1}(p-1)$; a generator is a <primitive root>.
Modulo seven, $3,3^2,\ldots,3^6$ are $3,2,6,4,5,1$, so three has order six. Moreover $3^6-1=728=7\cdot104$ is divisible by seven exactly once. We prove the lifting needed here. If $u=1+7^r v$ with $r\geq1$ 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 gives
$$
v_7(3^{6\cdot7^j}-1)=j+1.
$$
Any exponent giving one modulo $7^n$ is divisible by six. The order also divides $6\cdot7^{n-1}$ by <Lagrange theorem>, while the displayed valuation rules out each proper divisor of this form. Consequently \b[three is a <primitive root> modulo every $7^n$].
Since $2^3=1\pmod7$ 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 \b[two is never a <primitive root> modulo $7^n$]. Finally every unit modulo eight is odd, and $(2k+1)^2=1+4k(k+1)\equiv1\pmod8$. All unit orders are at most two, whereas the unit group has four elements. \b[There is no <primitive root> modulo eight.]
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