= Solution
The <direct product of groups> uses componentwise multiplication:
$$
(g,h)(g',h')=(gg',hh'),\qquad e=(e_G,e_H),\qquad
(g,h)^{-1}=(g^{-1},h^{-1}).
$$
Associativity follows in each factor. The maps $g\mapsto(g,e_H)$ and $h\mapsto(e_G,h)$ are injective <group homomorphisms> onto <subgroups>, giving the required copies of $G$ and $H$.
Use the PDF's convention that the subscript in $D_n$ is the <group> order. Write the hexagon <group> as $D_{12}=\langle r,s\mid r^6=s^2=e,\ srs=r^{-1}\rangle$ and the triangle <group> as $D_6=\langle a,b\mid a^3=b^2=e,\ bab=a^{-1}\rangle$. Let $z$ generate $C_2$. In $D_{12}$, $r^2$ has order three and $s$ conjugates it to its inverse, so $\langle r^2,s\rangle$ is a copy of $D_6$. The element $r^3$ has order two and is central, since $sr^3s=r^{-3}=r^3$.
Define
$$
\Phi(a^ib^j,z^k)=r^{2i+3k}s^j,
\qquad 0\le i<3,\quad 0\le j,k<2.
$$
The generator relations are preserved and $r^3$ commutes with the other images, so $\Phi$ is a <group homomorphism> from the direct product. The six residues $2i+3k$ are distinct modulo six: reduction modulo two first determines $k$, then reduction modulo three determines $i$. Together with $j$, they give all twelve distinct normal forms $r^ms^j$ in $D_{12}$. Thus $\Phi$ is bijective and
$$
\boxed{D_{12}\cong D_6\times C_2.}
$$
This is the <dihedral splitting when the half-rotation order is odd> with half-rotation order three.
Finally, $D_{12}$ contains an element of order six, namely $r$. The even cycle types on four symbols are the identity, a 3-cycle, and two disjoint transpositions; their orders are respectively one, three and two. Therefore $A_4$ has no element of order six. Since a <group isomorphism> preserves element orders,
$$
\boxed{D_{12}\not\cong A_4.}
$$
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