= Solution
Let $\varepsilon=1+\sqrt2$ and $P=(3+\sqrt2)$. The generator of this <principal ideal> has norm
$$
N_{K/\mathbb Q}(3+\sqrt2)=9-2=7.
$$
Reduction modulo $P$ gives $\sqrt2\equiv-3\equiv4\pmod7$. The map below is surjective and has $P$ in its kernel; both the kernel and $P$ have index seven, the latter by the <ideal norm>, so they are equal. Hence it induces an isomorphism of <residue fields>
$$
\mathcal O_K/P\cong\mathbb F_7,
\qquad
a+b\sqrt2\longmapsto a+4b\pmod7.
$$
In particular,
$$
\varepsilon\equiv5\pmod P.
$$
In the <finite field> $\mathbb F_7$, the element $5$ has multiplicative order six, since
$$
5^2\equiv4,
\qquad
5^3\equiv-1,
\qquad
5^6\equiv1\pmod7.
$$
By part (b), every unit is $(-1)^\delta\varepsilon^n$ for $\delta\in\{0,1\}$ and $n\in\mathbb Z$. Since $-1\equiv5^3$, its residue is $5^{n+3\delta}$, which equals one exactly when
$$
n+3\delta\equiv0\pmod6.
$$
The resulting units are $\varepsilon^{6k}$ and $-\varepsilon^{6k+3}$, and these are precisely the even and odd powers of $-\varepsilon^3$. Therefore
$$
\boxed{G=\langle-\varepsilon^3\rangle
=\left\langle-(1+\sqrt2)^3\right\rangle
=\langle-7-5\sqrt2\rangle},
$$
which is the group of <units congruent to one modulo 3 plus square root two>.
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