Answer to Question 1. Yes, $E(\mathbb{Z})=F$.

The inclusion $F\subseteq E(\mathbb{Z})$ follows from $E\subseteq E(\mathbb{Z})$ and Dickson's result $E=F$. It remains to show $E(\mathbb{Z})\subseteq F$. Pick $n\in E(\mathbb{Z})$. By definition there exist $u,v\in\mathbb{Z}^3$ such that $\lVert u\rVert=\lVert v\rVert$ and $n=\lVert u\rVert^2+\lvert u\cdot v\rvert$. Write $u=(a,b,c)$ and $v=(d,e,f)$. Then \begin{align*} 2n&=2\left(a^2+b^2+c^2+ad+be+cf\right)\\ &=a^2+b^2+c^2+2\left(ad+be+cf\right)+d^2+e^2+f^2\\ &=(a+d)^2+(b+e)^2+(c+f)^2 \end{align*} is a sum of three squares. The Theorem of Legendre implies that $2n$ cannot be written as $4^a\left(8b+7\right)$ with $a,b\geq 0$. From this we can conclude that $n$ must belong to $F$.

Answer to Question 1. Yes, $E(\mathbb{Z})=F$.

The inclusion $F\subseteq E(\mathbb{Z})$ follows from $E\subseteq E(\mathbb{Z})$ and Dickson's result $E=F$. It remains to show $E(\mathbb{Z})\subseteq F$. Pick $n\in E(\mathbb{Z})$. By definition there exist $u,v\in\mathbb{Z}^3$ such that $\lVert u\rVert=\lVert v\rVert$ and $n=\lVert u\rVert^2+\lvert u\cdot v\rvert$. Then $$2n =\lVert u\rVert^2+ \lVert v \rVert^2+ 2\lvert u\cdot v\rvert = \lVert u+v\rVert^2 $$ is a sum of three squares. Legendre's three-square theorem implies that $2n$ cannot be written as $4^a\left(8b+7\right)$ with $a,b\geq 0$. From this we can conclude that $n$ must belong to $F$.