Suppose that $f(x)$ and $g(x)$ are irreducible over $F$ and that $\deg f(x)$ and $\deg g(x)$ are relatively prime. 
Suppose that $f(x)$ and $g(x)$ are irreducible over $F$ and that $\gcd(~\deg g(x),\deg f(x)~)=1$. If $a$ is a zero of $f(x)$ in some extension of $F$, show that $g(x)$ is irreducible over $F(a)$

Attempt:  Let $b$ denote a zero of $g(x)$ in some extension of $F$. We have :
$[F(a,b):F(a)][F(a):F]=[F(a,b):F] = [F(a,b):F(b)][F(b):F] ~~~~..... (1)$
$\deg f(x) = [F(a):F ] ~~~~...... (2)$
$\deg g(x) = [F(b):F ] ~~~~......(3)$
Since, $\deg g(x)$ and $ \deg f(x)$ are prime to each other
$\implies [F(b):F]$ divides $[F(a,b):F(a)][F(a):F] \implies [F(b):F]$ divides $[F(a,b):F(a)]$
How do I move ahead?

Let $a,b \in \mathbb Q~~|~~b \neq 0$. Show that $\mathbb Q(\sqrt a) = \mathbb Q(\sqrt b)$ if and only if there exists some $c \in \mathbb Q$ such that $a = bc^2$

Attempt: Case $1$: When $ a = {p_1}^2/{q_1}^2$ where $p_1,q_1 \in \mathbb Z,\gcd(p_1,q_1)=1$
Then : $\mathbb Q(\sqrt a) = \mathbb Q(p_1/q_1) =\mathbb Q = \mathbb Q(\sqrt b)$ only if $\sqrt b = \dfrac {p_2}{q_2}$ or $b=\dfrac {{p_2}^2}{{q_2}^2} $
In such a case : $\dfrac {{p_1}^2}{{q_1}^2}=   \dfrac {{p_2}^2}{{q_2}^2} {(\dfrac {q_2 p_1}{p_2q_1})}^2$. Hence $c = (\dfrac {q_2 p_1}{p_2q_1})$
Did I do this correctly? How do i move ahead?
Thank you for your help..

You have $[F(b):F]\mid [F(a,b):F(a)]$. Now bound it from the other direction,
$$[F(a,b):F(a)]\le [F(b):F].$$
Note this is true for arbitrary $F,a,b$. Hint: consider $b$'s minpoly over $F(a)$ and $F$; what's a relationship between them. Once the indices are squeezed, they must be equal. Now what?
For your second question, you did Case 2 wrong. The number $\sqrt{a}$ can never be transcendental over the rationals, since it's a zero of $x^2-a$! First do the $a=bc^2\Rightarrow(\circ)$ part (easy) and then you can focus on the other direction. What kinds of things can you deduce from $\Bbb Q(\sqrt{a})=\Bbb Q(\sqrt{b})$? Right off the bat you should get a couple of containments, which means you can write things certain ways..
