$\DeclareMathOperator{\Gal}{Gal}$$\DeclareMathOperator{\Frob}{Frob}$
This comes down to checking a few things. Let $\kappa(K), \kappa(F)$ be the residue fields of $K$ and $F$.
Let $\rho: \operatorname{Gal}(\overline{K}/K) \rightarrow \operatorname{GL}(V)$ be a continuous, finite dimensional representation of the Weil group. Suppose that the kernel of $\rho$ contains $\operatorname{Gal}(\overline{K}/F)$, so we have a well defined homomorphism $\overline{\rho}: \Gal(F/K) \rightarrow \operatorname{GL}(V)$. The inertia group $I_K$ is the kernel of the surjective homomorphism
$$\Gal(\overline{K}/K) \rightarrow \Gal(\kappa(K)^{\operatorname{sep}}/\kappa(K))$$
and the inertia group $I_{F/K}$ is the kernel of the surjective homomorphism
$$\Gal(F/K) \rightarrow \Gal(\kappa(F)/\kappa(K))$$
First, a given $\sigma \in \Gal(\overline{K}/K)$ induces the Frobenius on $\Gal(\kappa(K)^{\operatorname{sep}}/\kappa(K))$ if and only if its image in $\Gal(F/K)$ induces the Frobenius on $\Gal(\kappa(F)/\kappa(K))$.
Second, the image of $I_K$ in $\Gal(F/K)$ is equal to $I_{F/K}$. Thus
$$\{v \in V : \rho(\sigma)v = v \textrm{ for all } \sigma \in I_K\} = \{v \in V : \overline{\rho}(\sigma)v = v \textrm{ for all } \sigma \in I_{F/K}\}$$
or $\rho^{I_K} = \rho^{I_{F/K}}$.
One thing you'll learn very quickly when it comes to duality theorems in Galois cohomology that that there are about a million things called duality, all slightly different from each other.
First, let's be clear on the definitions. Let $K$ be a local field and $K^s$ its separable closure. Let $G_K=\text{Gal}(K^s/K)$ be its absolute Galois group.
For a finite $G_K$ module $M$, define $M^D=\hom_{ab}(M,\mu(K^s))$. Note that we are taking the morphisms to the Galois module of all roots of unity, not just the ones in $K$. It is in this setting only that we have a canonical isomorphism
$$H^1(G_K,M)\simeq \widehat{H^1(G_K,M^D)}$$
I claim this does not hold for finitely generated modules $M$. Let $M=\mathbb{Z}$ have trivial action. Then $M^D=\hom_{ab}(\mathbb{Z},\mu(K^s))\simeq\mu(K^s)$.
First, we compute $H^1(G_K,M)$. Since $G_K\simeq\varprojlim\text{Gal}(L/K)$, we have
$$H^1(G_K,M)\simeq \varinjlim H^1(\text{Gal}(L/K),\mathbb{Z})\simeq \varinjlim \hom(\text{Gal}(L/K),\mathbb{Z})\simeq \varinjlim 0=0.$$
Now we compute $H^1(G_K,M^D)$. Note that $M^D\simeq\mu(K_S)\simeq\varinjlim \mu_n(K_S)$. So by proposition 1.5.1 of Cohomology of Number Fields (or because direct limits are exact) we have
$$H^1(G_K,M^D)\simeq \varinjlim H^1(G_K,\mu_n(K^S))$$
By Kummer theory, we know that $H^1(G_K,\mu_n(K^S))\simeq K^\times/(K^\times)^n$. So we have
$$H^1(G_K,M^D)\simeq \varinjlim K^\times/(K^\times)^n\simeq \varinjlim K^\times \otimes (\mathbb{Z}/n\mathbb{Z})\simeq K^\times \otimes \mathbb{Q}/\mathbb{Z}.$$
Therefore $H^1(G_K,M)\not \simeq \widehat{H^1(G_K,M^D)}$.
There is, however, a duality theorem that does apply to finitely generated modules. For a discrete finitely generated $G_K$ module $M$. In this case, we define $M^d=\hom_{ab}(M,(K^s)^\times)$ (Note that Milne also uses $M^D$ here, but I am using $M^d$ for clarity). This case becomes more complicated because some topology enters the picture, but we obtain isomorphisms
$$H^1(G_K,M)\simeq \widehat{H^1(G_K,M^d)}.$$
Note that this recovers the duality for finite modules as a special case: Let $M$ be a finite $G_K$ module. Since a homomorphism cannot take a finite order element to an infinite order element, we have
$$M^d=\hom_{ab}(M,(K^s)^\times)=\hom_{ab}(M,\mu(K^s))=M^D.$$
Best Answer
The fun of it is the Barsotti-Weil theorem, which says that for any abelian variety $A$ over $K$ we have $\widehat{A}(K) = \text{Ext}^1(A, \mathbb{G}_m)$, where $\widehat{A}$ is the dual abelian variety (elliptic curves are self-dual, of course). It's easy to say what the map between these groups is: a point of $\widehat{A}(K)$ is a degree $0$ line bundle on $A$, and removing the zero-section from such a bundle gives an extension of the group $A$ by $\mathbb{G}_m$. However, I don't know the proof that this map is an isomorphism.
Even spotting this, the work isn't done, since you need to identify extensions with Galois cohomology. Milne does this in Chapter I.3 of Arithmetic Duality Theorems, although he doesn't prove the Barsotti-Weil theorem itself.