Let $G$ be a linear algebraic group over a non-archimedean local field $F$. Let $H^1(F,G)$ be the first non-abelian Galois cohomology. It is known that when $F$ is of characteristic 0, i.e. finite extension of $Q\_p$, $H^1(F,G)$ is finite.


When $F=\mathbb{F\_q}((t))$, it is known that if $G$ is connected reductive group, then 
$H^1(F,G)$ is also finite. Now my question is that, for general $G$, what is the cardinality of $H^1(F,G)$? Is it always countable?



The cardinality of first non-abelian Galois cohomology
Since the fppf cohomology group ${\rm{H}}^1(F, \alpha\_p) = F/F^p$ is visibly uncountable (where $p = {\rm{char}}(F) > 0$), perhaps you meant to assume $G$ is smooth (and then fppf cohomology coincides with etale cohomology, which in turn coincides with Galois cohomology as in the title of the question). The cohomology set can be infinite even when $G$ is (smooth and) connected and commutative: see Example 11.3.3 in the book "Pseudo-reductive groups". 


But it is always *countable* for smooth $G$. Indeed, any right $G$-torsor $E$ over $F$ is $F$-smooth (since $G$ is) and hence is split by a finite separable extension $F'/F$ (by the Zariski-local structure theorem for smooth schemes over a field, or by more hands-on means), so it suffices to show that there are only countably many such $F'/F$ and that for each such $F'$ the set ${\rm{H}}^1(F'/F,G)$ of (isom. classes of) right $G$-torsors split by $F'/F$ is countable. 


In fact, each ${\rm{H}}^1(F'/F,G)$ is *finite* (so the infinitude for smooth affine $G$ is really caused by lack of control on splitting fields of torsors), but this is very hard to prove in general (see below), so let me first prove it is countable by more elementary means.


To check that $F$ has only countably many separable extensions $F'$ of degree below any specific bound we can apply the usual Krasner argument (as the space of *separable* Eisenstein polynomials of a given degree, while generally non-compact, has a countable base of open sets).


Now fix a finite separable extension $F'/F$ and consider the inclusion $G \hookrightarrow \mathscr{G} := {\rm{R}}\_{F'/F}(G\_{F'})$ of $F$-groups (using Weil restriction from $F'$ down to $F$). We are interested in the kernel of the induced map on ${\rm{H}}^1$'s (as this is identified with the restriction map ${\rm{H}}^1(F,G) \rightarrow {\rm{H}}^1(F',G)$ due to the non-abelian Shapiro Lemma). By Corollary 1 to Prop. 36 in section 5.4 of Chapter I of Serre's "Galois cohomology" book, this kernel is identified with the quotient set $\mathscr{G}(F)\backslash X(F)$ where $X$ is the smooth coset space $\mathscr{G}/G$ equipped with its natural left $\mathscr{G}$-action. But the natural quotient mapping $\mathscr{G} \rightarrow X$ is a smooth morphism (hence surjective on tangent spaces at $F$-points of the source), so likewise for the orbit map $\mathscr{G} \rightarrow X$ through *any* $x\_0 \in X(F)$. Hence, by the $F$-analytic implicit function theorem it follows that the induced map on $F$-points $\mathscr{G}(F) \rightarrow X(F)$ defined by $g \mapsto g.x\_0$ has open image, so all $\mathscr{G}(F)$-orbits in $X(F)$ are open. Since $X(F)$ is a countable base for its topology, we conclude that the space $\mathscr{G}(F)\backslash X(F)$ of $\mathscr{G}(F)$-orbits in $X(F)$ is countable.


The preceding countability proof was of elementary nature. To prove that each ${\rm{H}}^1(F'/F,G)$ is *finite*, we first note that this is elementary if $G$ is finite (etale). Indeed, it is harmless to increase $F'$ to be a Galois extension that splits $G$, and then this H$^1$ coincides with ${\rm{H}}^1({\rm{Gal}}(F'/F),G(F'))$ where both the Galois group and the coefficient group are finite, so finiteness is clear in such cases. In general, we may and do increase $F'$ so that it splits the finite etale $G/G^0$ and so that the natural map $G(F') \rightarrow (G/G^0)(F')$ is surjective. Thus, we have an exact sequence of pointed sets 
$${\rm{H}}^1(F'/F,G^0) \rightarrow {\rm{H}}^1(F'/F,G) \rightarrow {\rm{H}}^1(F'/F,G/G^0),$$
where the H$^1$'s in this diagram are for the Galois group ${\rm{Gal}}(F'/F)$ with coefficients in the groups of $F'$-points of $G^0$, $G$, and $G/G^0$ respectively.
By the known finiteness of the final term, it suffices to prove finiteness of the fiber through each point of the middle term. 


By the "twisting" method in Galois cohomology and the canonicity of the identity component (including its compatibility with ground field extension), each fiber is identified with an analogous "kernel" at the cost of replacing $G$ with an $F'/F$-form. In this way, the proof of finiteness of ${\rm{H}}^1(F'/F,G)$ is reduced to the case when the smooth $G$ is *connected*. As the OP noted, in the connected reductive case the finiteness is known (as even ${\rm{H}}^1(F,G)$ is finite in such cases, ultimately by deep results of Bruhat-Tits to prove vanishing in the simply connected semisimple case, with the general connected reductive case reducing to this via finiteness of $n$-torsion in Brauer groups of local fields and finiteness of degree-1 Galois cohomology of tori, which in turn rests on duality theorems and local class field theory). 


In the general smooth connected affine case one has to use the structure theory of pseudo-reductive groups to reduce the problem separately to the connected reductive case over finite (possibly inseparable) extensions of $F$ and to the connected solvable case (where one has to use Tits' structure theory of wound unipotent groups to treat the unipotent case). This final part of the argument is given in section 7.1 of the paper "Finiteness theorems for algebraic groups over function fields" in Compositio Math. 148 (2012) (see Prop. 7.1.2). In that section one also finds a finiteness result generalizing the stronger result in the connected reductive case, namely that if $G$ is pseudo-reductive and generated by its maximal $F$-tori (e.g., pseudo-reductive and perfect) then ${\rm{H}}^1(F,G)$ is finite.


