$K \lhd G, \space \space G/K \simeq H_1$, and $K \simeq H_2$
Solution 1:
$G = H_1 \times H_2$
Consider a map $\phi : H_1 \times H_2 \to H_1$ defined by
$$\phi (h_1, h_2) =h_1$$
Then $\phi$ is a onto homomorphism (check!) from $H_1 \times H_2 \to H_1$.
\begin{align}Ker(\phi) &=\{(h_1, h_2) :\phi(h_1, h_2)=e_{H_1}\}\\ &=\{(e_{H_1},h):h\in H_2\} \\ &=K\end{align}
Since,$\phi : H_1 \times H_2 \to H_1$ is an onto homomorphism with $Ker(\phi) =K$.
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Hence, $K$ is a normal subgroup of $G= H_1 \times H_2$
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Then, by first Isomorphism Theorem, $G/K \simeq H_1$
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Define a map $\mu:K \to H_2$ by
$\mu(e_{H_1}, h) =h $ for all $h\in H_2$
This gives an isomorphism from $K$ to $H_2$.