If $H$ is a normal subgroup of $G$, is $G/H \times H \cong G$?

Let me add a couple of comments to the other excellent answers.

In some cases there is a subgroup $K$ of $G$ such that $G = H K$ and $H \cap K = 1$. Then $G/H \cong K$, and $G$ is a semidirect product (a.k.a. split extension) $H \rtimes K$ of $H$ by $K$. The semidirect product can be regarded as a generalization of the direct product, in which only one of the factors is normal.

In general, even such a subgroup $K$ of $G$ will not exist. Define then $K = G/H$. Now $G$ is said to be an extension of $H$ by $K$. Things get immediately very difficult here. However, one can define a section as a map $\sigma : K \to G$ such that $\varphi \circ \sigma = \textbf{1}_{K}$. (Here $\varphi : G \to K = G/H$ is the canonical map.) $\sigma(K)$ will not be a subgroup of $G$ (unless $G$ is a split extension of $H$ by $K$ as above), but the multiplication between elements of $\sigma(K)$ will have to be corrected by an element of $H$. This yields what is called a $1$-cocycle, and from now on you are in the territory of group cohomology.

No. If $A\times B\cong G$ then both $A$ and $B$ (respectively their images under the isomorphism) are normal in $G$. Your example with $\mathbb Z/2\mathbb Z$ does not work because $\mathbb Z/2\mathbb Z$ is not normal in $\mathbb Z$, yes it is not even a subgroup (no nonzero element $x\in \mathbb Z$ has the property $x+x=0$).

Any cyclic group whose order is not squarefree contains at least one normal subgroup for which this fails, as $C_{p^2m}\not\cong C_{p}\times C_{pm}$.

You may be looking for the Schur-Zassenhaus theorem.