Factor $f(x)$ into irreducibles in $K[x]$,
$$f(x) = q_1(x)\cdots q_m(x).$$
If $\alpha$ is a root of $q_1(x)$ and $\beta$ is a root of $q_i(x)$, then $q_1(x)$ is the irreducible polynomial of $\alpha$ over $K$, and $q_i(x)$ is the irreducible polynomial of $\beta$ over $K$.
Since the action of $G$ is transitive on the roots of $f$ there exists $\sigma\in G$ such that $\sigma(\alpha)=\beta$. Since $K$ is Galois, $\sigma(K)=K$, so $\sigma(q_1(x))\in K[x]$ is a polynomial that has $\sigma(\alpha)=\beta$ as a root. Therefore, $q_i(x)|\sigma(q_1(x))$. Since both are irreducible, it follows that $\deg(q_i) = \deg(\sigma(q_1))=\deg(q_1)$. So all irreducible factors of $f(x)$ in $K$ have the same degree.
The degree is equal to the degree of the extension $K(\alpha)/K$ (which is the degree of $q_1(x)$).
A for $m$, the number of factors, we have:
$n=[F(\alpha):F] = [F(\alpha):F(\alpha)\cap K][F(\alpha)\cap K:F]$.
I claim that $[F(\alpha):F(\alpha)\cap K] = [K(\alpha):K]$.
Indeed, since $K$ is Galois over $F$, then $K(\alpha)$ is Galois over $F(\alpha)$ ($K$ is the splitting field of some polynomial over $F$, and this same polynomial works for $K(\alpha)$), and $K$ is Galois over $K\cap F(\alpha)$. If $\sigma\in\mathrm{Gal}(K(\alpha)/F(\alpha))$, then restricting $\sigma$ to $K$ gives a homomorphism $\mathrm{Gal}(K(\alpha)/F(\alpha))\to \mathrm{Gal}(K/K\cap F(\alpha))$. If $\sigma$ restricts to the identity on $K$, then it must be the identity on $K(\alpha)$ (it fixes $\alpha$ since it fixes $F(\alpha)$ pointwise), so the map $\mathrm{Gal}(K(\alpha)/F(\alpha))\to \mathrm{Gal}(K/K\cap F(\alpha))$ is one-to-one.
Let $H'$ be the image of this map. Then $H'$ fixes $K\cap F(\alpha)$ pointwise, and if $k\in K$ is fixed by all elements of $H'$, then $k$ must be fixed by all elements of $\mathrm{Gal}(K(\alpha)/F(\alpha))$, hence $k\in F(\alpha)\cap K$. So $F(\alpha)\cap K$ is the fixed field of $H'$, hence $H'=\mathrm{Gal}(K/K\cap F(\alpha))$.
Thus, $[K(\alpha):F(\alpha)] = [K:K\cap F(\alpha)]$.
Now,
$$\begin{align*}
[K(\alpha):K\cap F(\alpha)] &= [K(\alpha):K][K:K\cap F(\alpha)]\\
\text{ and }[K(\alpha):K\cap F(\alpha)] &= [K(\alpha):F(\alpha)][F(\alpha):K\cap F(\alpha)].\end{align*}$$
Therefore, $[K(\alpha):K] = [F(\alpha):K\cap F(\alpha)]$, as claimed.
So we have that
$$\begin{align*}
n &= [F(\alpha):F]= [F(\alpha):F(\alpha)\cap K][F(\alpha)\cap K:F]\\
&= [K(\alpha):K][F(\alpha)\cap K:F] \\
&= d[F(\alpha)\cap K:F].
\end{align*}$$
Since $n=dm$, then
$$m = [F(\alpha)\cap K:F],$$
as claimed.
To see the action of $G$ is transitive, note that $F(\alpha)$ is isomorphic to $F(\beta)$ over $F$, since $\alpha$ and $\beta$ have the same irreducible over $F$; hence there is an isomorphism $\sigma\colon F(\alpha)\to F(\beta)$ that restricts to the identity on $F$ and sends $\alpha$ to $\beta$. Since $L$ is Galois over $F$, and $F(\alpha),F(\beta)\subseteq L$, $\sigma$ extends to an element of $G$.
First, note that the first part of the question implies the second part: i.e. suppose $f$ has roots in $L$. Then, $f$ factors into product of irreducible factors with same degree, one of which has degree 1 (since $f$ has a root in $L$). Therefore, $f$ splits over $L$.
For the first part, take $\bar{K}$ to be the algebraic closure of $K$. And consider $\sigma\in \textrm{Gal}(\bar{K},K)$ where $\sigma$ permutes the roots of $f$. Now, suppose $f$ factors as the product of $f_1\cdots f_n$ over $L$. Can you use the fact that $L/K$ is normal to show $\sigma|_L$ sends each $f_i$ to some $f_j$? (we are not ruling out the case $i=j$)
After that, you can choose appropriate $\sigma$ to show that you can actually send $f_1$ to $f_j$ for any $j$.
I hope this helps!
Best Answer
If $f=pq$ where $p$ is irreducible in $K[x]$ with degree $r$ and $0＜r＜n$.select a root $\alpha$ of $p$ .then $[K[\alpha]:K][K:F]=[K[\alpha]:F[\alpha]][F[\alpha]:F]$,that is $rm=[K[\alpha]:F[\alpha]]n$.Hence $n\mid r$ since n and m are coprime. Contradiction