= Solution
The element $\omega$ satisfies
$$
\omega^2-\omega+12=0.
$$
Modulo both $2$ and $3$, this polynomial factors as $X(X-1)$. Hence
$$
(2)=\mathfrak p_2\mathfrak q_2,
\quad
\mathfrak p_2=(2,\omega),
\quad
\mathfrak q_2=(2,\omega-1),
$$
and
$$
(3)=\mathfrak p_3\mathfrak q_3,
\quad
\mathfrak p_3=(3,\omega),
\quad
\mathfrak q_3=(3,\omega-1),
$$
where the respective <ideal norms> are $2,2,3,3$.
Now $N(\omega)=12$, and $\omega$ lies in $\mathfrak p_2$ and $\mathfrak p_3$ but not in their conjugate primes. Similarly $N(2+\omega)=18$, while $2+\omega$ lies in $\mathfrak p_2$ and $\mathfrak q_3$. Comparing norms gives
$$
\boxed{(\omega)=\mathfrak p_2^2\mathfrak p_3,
\qquad
(2+\omega)=\mathfrak p_2\mathfrak q_3^2.}
$$
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