Is every normal subgroup of $G$ of this form? 
Let $G = H \times K$. If $N$ is a normal subgroup of $H$ and $L$ is a normal subgroup
of $K$, show that $N \times L$ is a normal subgroup of $G$. Is every normal subgroup of $G$ of
this form?

My attempt:

Let $G = H \times K$. If $N$ is a normal subgroup of $H$ and $L$ is a normal subgroup
of $K$, show that $N \times L$ is a normal subgroup of $G$

Since $e\in N$ and $e\in L$, then $(e,e)\in N\times L$. If $(n_1,l_1),(n_2,l_2) \in N\times L$, then $(n_1,l_1)(n_2,l_2)=(n_1n_2,l_1l_2)\in N\times L$. Also, if $n\in N$, then $n^{-1}\in N$ and if $l\in L$, then $l^{-1}\in L$, so $(n,l)(n^{-1},l^{-1})=(nn^{-1},ll^{-1})=(e,e)$. Thus $N\times L$ has inverses and therefore, $N\times L \leq G$.
For normality, if $(h,k)$ is an element of $G$ and $(n,l)\in N\times L$, then $(h^{-1},k^{-1})(n,l)(h,k)=(h^{-1}nh,k^{-1}lk)$. Since $N$ is normal in $H$ and $L$ is normal in $K$, we have $h^{-1}nh\in N$ and $k^{-1}lk\in L$, so $(h^{-1}nh,k^{-1}lk) \in N\times L $ and therefore $N\times L$ is a normal subgroup of $G$.

Is every normal subgroup of $G$ of this form?

In this part, my intuition says yes, but I don't know how to formalize this, could you help me to formalize the answer, either positive or negative?

No. Consider for example the group $G=\Bbb Z/2\Bbb Z\times \Bbb Z/2\Bbb Z$. It has the (normal) subgroup $\{(0,0),(1,1)\}$ which is not of the desired form. However under certain conditions it is true, for example when $H,K$ are finite of coprime order.
