1. ## Algebraic

let a = sqrt(7+4*sqrt(3)) prove that a is algebraic over Q

now let a = sqrt(7) + i show that a is now algebraic over R

2. ## Polynomials

You just gotta find a polynomial over the respective field such that a is a root. Just takes some playing around with the numbers.

1)
$\displaystyle a^2=7+4\sqrt3$
$\displaystyle a^4=(a^2)^2=97 + 56\sqrt3$

so $\displaystyle p(x) = x^4-14x^2+1 \in \mathbb{Q}[x]$ has a as a root. Dunno if it is irreducible or not, but you didn't ask for the degree of the extension. Clearly either 2 or 4 depending on whether p(x) factors into square terms. (Rational root test shows no linear terms).

2)We are in $\displaystyle \mathbb{R}$ now, so in particular, $\displaystyle \sqrt{7} \in \mathbb{R}$.

Consider $\displaystyle (a-\sqrt7)^2+1=0$ so a is a root to $\displaystyle (x-\sqrt7)^2+1=0$ This pretty much has to be minimal as i is in complex and is a degree 2 extension over the reals, so it should be a degree 2 extension over the reals.

Hope this helped