= Solution
Use the given square-class assignments and $\omega^3=1$, $1+\omega+\omega^2=0$. The <character> values, in the specified class order, are
$$
\boxed{\chi_S=(3,3,-1,0,0,0,0),\qquad\chi_A=(1,1,1,\omega,\omega^2,\omega^2,\omega).}
$$
For example the fourth entry of $\chi_S$ is $(( -\omega^2)^2-\omega)/2=0$, and that of $\chi_A$ is $\omega$. A class of centralizer order $c$ has size $24/c$, so a <character> norm is the sum of its squared moduli divided by these centralizer orders. Therefore
$$
\langle\chi_S,\chi_S\rangle=\frac9{24}+\frac9{24}+\frac14=1,\qquad
\langle\chi_A,\chi_A\rangle=\frac1{24}+\frac1{24}+\frac14+\frac46=1.
$$
Both are genuine <characters> by their symmetric/exterior-square construction. A <character> norm is the sum of squares of its irreducible multiplicities, so \b[both <characters> are irreducible].
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