Your logic going from $f(a)=g(\bar{a})$ and concluding that they must have the same roots is flawed. How do you know that? The way to prove the first part is to take the minimal polynomial of $a$, $f(x)$, and showing that
$$
0=f(a)=\overline{f(a)}=f(\overline{a})\ ,
$$
and concluding that $f$ is also the minimal polynomial of $\overline{a}$. You can go figure out the details, in particular, why we can say $\overline{f(a)}=f(\overline{a})$ and why $f(\overline{a})$ is not only a polynomial with $\overline{a}$ as a root, but must also be its minimal polynomial.
EDIT: My original answer to the second part was completely wrong. Here is a counterexample which shows that the answer to part $2$ of the question is 'No'.
Consider the polynomial $f(X)=X^3-2$ over $\mathbb{Q}$. This is the go-to example for a non-normal extension of $\mathbb{Q}$. Let $\alpha=\sqrt[3]{2}$ and $\omega=e^{2\pi i/3}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$, then the roots of $f$ are $\alpha, \omega\alpha$ and $\omega^2\alpha$. It can be shown that $\mathbb{Q}(\alpha)$, $\mathbb{Q}(\omega\alpha)$ and $\mathbb{Q}(\omega^2\alpha)$ are all isomorphic and that all their degrees over $\mathbb{Q}$ are $3$. Taking $a=\omega\alpha$ gives the desired result.
We can simplify the expression $\alpha^3 - 14 \alpha$. This occurs because $\alpha^2$ satisfies $\alpha^4 = 16\alpha^2 - 4$, therefore $$\alpha^3 = 16\alpha - 4\alpha^{-1} $$
and so $\theta = \alpha^3 - 14\alpha = 2\alpha - 4\alpha^{-1}$. (note that $\alpha \neq 0$ so is invertible)
Let us square this term :
$$
\theta^2 = 4\alpha^2 - 16\alpha^{-2} -16 = \frac{4\alpha^4 - 16}{\alpha^2} - 16 = 4\frac{\alpha^4 - 4}{\alpha^2} -16
$$
but $\alpha^4 - 4 = 16 \alpha^2$! So substitute that in and get $4 \times 16 - 16 = 48 = \theta^2$.
Thus, the minimal polynomial should be $x^2-48$ as long as you see that is irreducible.
We are lucky that such a manipulation worked out. In general, however, we'd either have to visit higher powers, or need good manipulative intuition to find the minimal polynomial. In some cases, knowing a root can be helpful, so you can directly compute the expression for that root.
For example, the polynomial given is in fact the minimal polynomial of $\sqrt 3 + \sqrt 5$, so one can try the manipulation with this as $\alpha$ to see that $\alpha^3 - 14 \alpha = 4\sqrt 3$ (in case that no intelligent techniques work, this is brute force).
Best Answer
Consider the minimal polynomial of $a_{i+1}$ over $\Bbb{Q}[a_1,...,a_i]$. It divides $x^2-a_{i+1}^2\in\Bbb{Q}[x]$ and so must have degree $1$ or $2$. But this means the degree of $\Bbb{Q}[a_1,...,a_i,a_{i+1}]$ over $\Bbb{Q}[a_1,...,a_i]$ is $1$ or $2$. Thus, if the degree of $\Bbb{Q}[a_1,...,a_i]$ over $\Bbb{Q}$ is a power of $2$, say $2^j$, then the degree of $\Bbb{Q}[a_1,...,a_i,a_{i+1}]$ over $\Bbb{Q}$ is the product of the degree of $\Bbb{Q}[a_1,...,a_i,a_{i+1}]$ over $\Bbb{Q}[a_1,...,a_i]$ and that of $\Bbb{Q}[a_1,...,a_i]$ over $\Bbb{Q}$, and is thus $2^j$ or $2^{j+1}$, in either case a power of $2$. By induction, then, the degree of $\Bbb{Q}[a_1,...,a_n]$ over $\Bbb{Q}$is a power of $2$.