First, suppose that $L/K$ is separable. Fix an algebraic closure $\overline{K}$ for $K$, and let $M$ be a Galois closure for $L/K$ in $\overline{K}$.
EDIT: If the idea of Galois closure isn't known to you, think about it as follows. Since your extension is separable, it is primitive, say $L=K(\alpha)$. Note then that necessarily $K(\alpha')/K$ is separable for every root $\alpha'$ of the minimal polynomial $m_\alpha$ of $\alpha$ over $K$. In the algebraic closure $\overline{K}$, all the roots $\alpha'$ exist. Then, the splitting field $F$ of $m_\alpha$ is just $K(\{\alpha':m_\alpha(\alpha')=0\})$. Since each $\alpha'$ is separable the full extension $F$ is separable. What this basically shows is that every finite separable extension of $K$ is contained in a finite Galois extension of $K$ (in fact, if you fix an algebraic extension, this is the minimal Galois extension containing $K$). So, when I say "take a Galois closure" I just mean "think about a Galois extension containing our extension". This is a common technique, for it allows you to deduce results about separable extensions by first proving it for Galois extensions, and then deducing it for certain subextensions.
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Note that for any embedding $\tau\in\text{Hom}_K(L,\overline{K})$ one necessarily has that $\tau(L)\subseteq M$. Thus, for every $\sigma\in\text{Gal}(M/K)$ and every $\tau\in\text{Hom}_K(L,\overline{K})$ one has that $\sigma\circ\tau\in\text{Hom}_K(L,\overline{K})$. Moreover, since $\sigma\circ\tau=\sigma\circ\tau'$ implies that $\tau=\tau'$, you can see, in particular, that the mapping $\tau\mapsto \sigma\circ\tau$ is an injection $\text{Hom}_K(L,\overline{K})$ to itself, and so necessarily a bijection. From this you can see that for every $\sigma\in\text{Gal}(M/K)$ and $x\in L$ one has that
$$\begin{aligned}\sigma\left(\text{Tr}_{L/K}(x)\right) &= \sigma\left(\sum_{\tau\in\text{Hom}_K(L,\overline{K})}\tau(x)\right)\\ &=\sum_{\tau\in\text{Hom}_K(L,\overline{K})}(\sigma\circ\tau)(x)=\\ &=\sum_{\tau\in\text{Hom}_K(L,\overline{K})}\tau(x)\\ &= \text{Tr}_{L/K}(x)\end{aligned}$$
Since $M/K$ is Galois, this implies that $\text{Tr}_{L/K}(x)\in K$.
Do something similar for the norm map.
For your second question. Use the fact that $F\subseteq F(\alpha)\subseteq K$ and the fact that degree is multiplicative in towers. To deduce the other result, consider the polynomial
$$\prod_{\sigma\in\text{Hom}_F(K,\overline{F})}(T-\sigma(\alpha))$$
Convince yourself that for each distinct Galois conjugate $\beta$ of $\alpha$, that $\sigma(\alpha)=\beta$ precisely $\frac{n}{d}$ times. Thus,
$$\prod_{\sigma\in\text{Hom}_F(K,\overline{F})}(T-\sigma(x))=m_\alpha(T)^{\frac{n}{d}}$$
Compare the constant term and the coefficient of $T^{n-1}$ on both sides to get both of your results.
We need the following lemma.
Lemma
Let $K/F$ be a (not necessarily finite dimensional) Galois extension,
$L/F$ an arbitrary extension.
Clearly $KL/L$ is Galois.
Then the restriction map, namely, $\sigma\mapsto \sigma\mid K$ induces
an isomorphism $\psi\colon \mathrm{Gal}(KL/L) \rightarrow \mathrm{Gal}(K/K\cap L)$.
Proof:
We regard $\mathrm{Gal}(KL/L)$ and $\operatorname{Gal}(K/K\cap L)$ as topological groups with Krull topologies.
Clearly $\psi$ is continuous and injective.
Let $H = \psi(\mathrm{Gal}(KL/L))$.
Since $\mathrm{Gal}(KL/L)$ is compact, $H$ is also compact.
Since $\mathrm{Gal}(K/K\cap L)$ is Hausdorff, $H$ is closed.
Clearly the fixed subfield of $K$ by $H$ is $K \cap L$.
Hence $H = \mathrm{Gal}(K/K\cap L)$ by the fundamental theorem of (not necessarily finite dimensional) Galois theory.
This completes the proof.
Now we prove the following proposition with which the OP had a problem.
Proposition
Let $K$ and $L$ be Galois extensions of $F$.
The restriction of function map, namely, $\sigma\mapsto(\sigma\vert_K,\sigma\vert_L)$ induces a group homomorphism $\varphi\colon\operatorname{Gal}(KL/F)\to\operatorname{Gal}(K/F)\times\operatorname{Gal}(L/F)$. Suppose $K\cap L=F$. Then $\varphi$ is an isomorphism.
Proof.
Since it is clear that $\varphi$ is injective, it suffices to prove that it is surjective.
Let $G_1 = \mathrm{Gal}(K/F), G_2 = \mathrm{Gal}(L/F), G = \mathrm{Gal}(KL/F)$.
By the lemma, given $\sigma_1 \in G_1$, there exists $\sigma \in \mathrm{Gal}(KL/L)$ such that $\sigma\mid K = \sigma_1$. Since $\sigma \in G$ and $\sigma\mid L = 1_L$, $G_1\times 1 \subset \varphi(G)$.
Similarly $1\times G_2 \subset \varphi(G)$.
Hence $G_1\times G_2 = \varphi(G)$.
This completes the proof.
Best Answer
$H = \text{Gal}(L/F)^K \subset H' = \text{Gal}(L/F)^{F(x)} \subset \text{Gal}(L/F)$
$n = [K:F] = [ \text{Gal}(L/F) : H]$, and $d = [K(x):F] = [\text{Gal}(L/F) : H']$, so $n/d = [H' : H]$
$N_{K/F}(x) = \prod_{\sigma H \in \text{Gal}(L/F) / H} \sigma(x)$ So you need to find out when two $H$-cosets induce the same image for $\sigma(x)$.
The image $\sigma(x)$ only depends on the $H'$-coset containing $\sigma$, since $x$ is fixed by $H'$. Now, since $H \subset H'$, an $H'$-coset is a reunion of $H$-cosets : $H' = \bigcup_{\tau H \in H'/H} \tau H$, so for every $\sigma H' \in \text{Gal}(L/F) / H'$, $\sigma H' = \bigcup_{\tau H \in H'/H} \sigma \tau H$
Every $H'$-coset contains exactly $n/d$ $H$-cosets, so when you group the factors accordingly,
$N_{K/F}(x) = \prod_{\sigma H' \in \text{Gal}(L/F) / H'} \prod_{\tau H \in H'/ H}\sigma\tau(x) = \prod_{\sigma H' \in \text{Gal}(L/F) / H'} (\sigma(x))^{n/d}$
which is precisely what you need to show.