It is pretty easy to show that given a prime $p$, $$A = \mathbb{Q}(i, \sqrt{p}) \cong \mathbb{Q}(i + \sqrt{p}) = B$$
Obviously $B \subseteq A$, and by manipulating $i + \sqrt{p}$, we can conclude that $$\{i,\sqrt{p}\} \subset B \implies A \subseteq B$$
But I'm just wondering if it is still true for a non-prime $p$?
Hint $\,\ u = \sqrt n + \sqrt m \in B$
$\ \ \ \Rightarrow\ v := \sqrt n - \sqrt m = \dfrac{n-m}{\sqrt n + \sqrt m} = \dfrac{n-m}u\in B$
Thus $\ u+v = 2\sqrt n\in B,\,$ so $\,\sqrt n\in B,\ $ so $\,\sqrt{m} = u - \sqrt n\in B$
Remark $ $ The key idea behind the proof is as follows. If field $\rm\,F\,$ has two $\rm\,F$-linear independent combinations of $\rm\, \sqrt{n},\ \sqrt{m}\, $ then we can solve for $\rm\, \sqrt{n},\ \sqrt{m}\, $ in $\,\rm F.\,$ In our case we notice $\rm\ F = \mathbb Q(\sqrt{n} + \sqrt{m})\ $ contains the independent $\rm\ \sqrt{n} - \sqrt{m}\ $ since
$$\rm \sqrt{n} - \sqrt{m}\ =\ \dfrac{\ a\,-\,b}{\sqrt{n}+\sqrt{m}}\ \in\ F = \mathbb Q(\sqrt{n}+\sqrt{m}) $$
To be explicit, notice that $\rm\ u = \sqrt{n}+\sqrt{m},\ v = \sqrt{n}-\sqrt{m}\in F\ $ so solving the linear system for the roots yields $\rm\ \sqrt{n}\ =\ (u+v)/2,\ \ \sqrt{m}\ =\ (u-v)/2,\ $ both of which are clearly $\rm\,\in F,\:$ since $\rm\:u,\:v\in F\:$ and $\rm\:2\ne 0\:$ in $\rm\:F,\:$ so $\rm\:1/2\:\in F.\:$ This works over any field where $\rm\:2\ne 0,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.
This idea often proves useful , e.g. the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see PlanetMath's proof.
$(\sqrt n+i)^3=n\sqrt n+3in-3\sqrt n-i=(n-3)\sqrt n+(3n-1)i$. $(\sqrt n+i)^3-(n-3)(\sqrt n+i)=(2n+2)i$. So, $i$ is in $B$, so $\sqrt n$ is in $B$, so $A=B$.
This doesn't work when $n=-1$, but that's not a problem.
This is true. As we have that containment, it suffices to show that the degree of $\mathbb{Q}(i, \sqrt{n})$ over $\mathbb{Q}$ is the same as the degree of $\mathbb{Q}(i+\sqrt{n})$ over $\mathbb{Q}$. It's easy to show that $[\mathbb{Q}(i, \sqrt{n}): \mathbb{Q}] = 4$. To show that the same is true for $\mathbb{Q}(i + \sqrt{n})$, we just need to find the degree of the minimal polynomial for $i + \sqrt{n}$ over $\mathbb{Q}$.
The degree of this polynomial will be equal to the total number of distinct Galois conjugates of $i + \sqrt{n}$ per this result (these will be its roots). One can check that the only valid $\mathbb{Q}$-automorphisms of this field are generated by $i \mapsto -i$ and $\sqrt{n} \mapsto - \sqrt{n}$, which yields a set of four Galois conjugates: $\pm i \pm \sqrt{n}$. Thus, $[\mathbb{Q}(i + \sqrt{n}): \mathbb{Q}] = 4$, and the two fields are isomorphic.