= Solution
Only the <character> $\chi_2$ of the two-dimensional <irreducible representation> remains unknown. Write it as $(2,a,b,c,d)$. The <regular representation> has <character> zero off the identity and decomposes with multiplicities equal to the <irreducible representation> dimensions. Thus, for a nonidentity element $g$,
$$
\chi_0(g)+\chi_1(g)+2\chi_2(g)+3\chi_3(g)+3\chi_4(g)=0.
$$
For the four nonidentity <conjugacy classes> this gives $a=0$, $b=2$, $c=-1$, $d=0$. The complete <character table> is consequently
$$
\boxed{
\begin{array}{c|rrrrr}
&e&(12)&(12)(34)&(123)&(1234)\\\hline
\text{class size}&1&6&3&8&6\\\hline
\chi_0&1&1&1&1&1\\
\chi_1&1&-1&1&1&-1\\
\chi_2&2&0&2&-1&0\\
\chi_3&3&1&-1&0&-1\\
\chi_4&3&-1&-1&0&1
\end{array}}
$$
For example, $\langle\chi_2,\chi_2\rangle=(4+12+8)/24=1$, agreeing with <character orthogonality>.
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