= Solution
Let
$$
K=\{e,(12)(34),(13)(24),(14)(23)\}.
$$
The product of any two distinct nonidentity elements is the third, and each squares to $e$, so $K$ is a <Klein four-group>. Conjugating a double transposition simply relabels its letters, keeping it in the set. Hence $K$ is a <normal subgroup> of $A_4$.
Choose $r=(123)$. The <cosets> $K,rK,r^2K$ are distinct and cover the twelve elements. With $\zeta=e^{2\pi i/3}$, define
$$
\boxed{\Phi(r^jk)=\zeta^j,\qquad j\in\{0,1,2\},\quad k\in K.}
$$
The coset decomposition makes this well-defined. Normality gives
$$
(r^jk)(r^\ell k')=r^{j+\ell}(r^{-\ell}kr^\ell)k'\in r^{j+\ell}K,
$$
so $\Phi$ preserves products. It is onto $\{1,\zeta,\zeta^2\}\cong C_3$, and
$$
\boxed{\ker\Phi=K,\qquad A_4/K\cong C_3.}
$$
This is the <tetrahedral quotient onto the cyclic group of order three>. In terms of (a), it sends $C_0\cup C_2$ to $1$, $C_+$ to $\zeta$, and $C_-$ to $\zeta^2$. The normality check above also follows abstractly because a <group homomorphism kernel> is normal.
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