@A.Stasinski: Sure, any example for which $G(R)$ is finite would do the job when $\dim G_F > 0$. For example, as recently came up in a comment to the question about geometric unirationality, let $R = k[\![t]\!]$ with $k$ algebraically closed of characteristic $p > 0$ and let $G$ be the affine flat $R$-group scheme given by the equation $y^q = x - t x^p$ for a $p$-power $q > 2$. Then $G_F$ is smooth and $G(F)$ consists of $\{(0,0)\}$ when $p > 2$ whereas it consists of $\{(0,0), (1/t,0)\}$ when $p=2$, so $G(R) = 0$ for any $p$ and hence $H_k=0$.

@Jason: Oops, I got myself confused for what is needed with $k$-points, sorry.

@A.Stasinski: The equality $\widehat{G}(R)=G(R)$ is the map induced on $R$-points by the $R$-morphism $f:\widehat{G}\rightarrow G$ (and $f_F$ is the identification of $F$-fibers, which identifies $F$-points too). So the commutativity just expresses the fact that the "functor of points" of a scheme is functorial in the scheme.

@ACL: What is a reference for this particular "Greenberg Theorem"? I wondered about trying to use the Greenberg functor to prove the constructibility, but when passing from artinian quotients of $R$ to the inverse limit $R$ itself I got stuck on issues of non-emptiness of a countable descending intersection when $k$ is countable.

@Jason: In char. 0 singularities are fine since a resolution cannot have $k$-points if the original variety doesn't. Also, $G(k)\rightarrow(G/P)(k)$ is surjective for any connected reductive $k$-group $G$ and parabolic $k$-subgroup $P$ over any $k$, so ${\rm{H}}^1(k,P)\rightarrow{\rm{H}}^1(k,G)$ has trivial kernel. Hence, any nontrivial $B$-torsor over $k$ does the job in char. 0. If $\tilde{G}$ is the quasi-split simply connected central cover and $\tilde{T}$ a maximal $k$-torus in a Borel, it suffices to find a $k$-subgroup $\mu\subset Z_{\tilde{G}}$ so ${\rm{H}}^1(k,\tilde{T}/\mu)\ne 1$.

@Jason: is there a literature reference for the wonderful compactification over a general (infinite) field? All references I have seen assume the field to be algebraically closed. Also, to have a nontrivial $B$-torsor over $k$ we cannot take $G$ to be either simply connected or adjoint, since in such cases ${\rm{H}}^1(k,B)=1$ (as the maximal $k$-tori of $B$ are "induced", and hence have vanishing ${\rm{H}}^1$, in these two extreme cases). So what sort of $G$ do you plan to try?

The question was revised to require $V$ to be projective and smooth, precisely to avoid such examples (noted in the comments to the question by ayanta).

If $\chi_1,\dots,\chi_r$ are the characters of the fundamental representations then $\chi$ is the map $g \mapsto (\chi_i(g))$, so why don't Theorem 8.1 and Lemma 8.5 in Steinberg's IHES paper answer the 2nd question? If you write a regular semisimple element as an explicit conjugate of a regular element of $T$ then the problem seems to "reduce" to the case of regular elements of $T$, and $(G/T) \times T \rightarrow G$ defined by $(g,t)\mapsto gtg^{-1}$ is finite etale over the Zariski-open regular semisimple locus, so it would help if you define what you mean by "compute"...in terms of what?

What does "any normal subgroup $N$" mean? Not every subgroup sheaf of the functor represented by $G$ is represented by an affine algebraic group (e.g., can make $N$ representable not by an affine group scheme). You meant to ask: for a closed normal subgroup scheme $N$ of $G$, is there an fppf (hence fpqc!) homomorphism of affine algebraic groups $G\rightarrow Q$ with kernel $N$, and if so then is $G\rightarrow Q$ also etale? The 1st is affirmative over any field (use generic flatness and geometric translations), the 2nd is an exercise in fppf descent theory since $N$ is smooth (char. 0!).

An easy case for central division algebras is $p=2$: a smooth conic over $K$ with no $K$-point. Consider $ux^2=y^2+tyz+tz^2$ in $\mathbf{P}^2_K$ for $u\in R^{\times}$ with reduction in $k$ not a square and $t$ a uniformizer. (Allow mixed-characteristic $K$, as in the answer below.) If $[x_0,y_0,z_0]$ is a $K$-point on this conic then WLOG $x_0,y_0,z_0\in R$ with at least one of them a unit. Clearly $z_0$ must be a unit, so WLOG $z_0=1$, and hence $ux_0^2=y_0^2+ty_0+t$. Then ${\rm{ord}}(y_0)={\rm{ord}}(x_0)\le 0$, so we get a contradiction since $u$ isn't residually a square.

Do you also want a simply connected absolutely simple connected semisimple group over $K = k((t))$ with non-vanishing $H^1$ whenever $k$ is separably but not algebraically closed? If $p := {\rm{char}}(k) > 0$ then there are Galois extensions $K'/K$ of degree $p$, and the norm ${K'}^{\times} \rightarrow K^{\times}$ is not surjective, so there are central division algebras $D$ over $K$ with dimension $p^2$ containing $K'$ as a maximal commutative subalgebra, and $H^1(K, {\rm{SL}}_1(D)) \ne 1$ provided the reduced norm $D^{\times}\rightarrow K^{\times}$ is not surjective. Do you want such $D$?

Presumably $T$ is meant to be a torus over $k((t))$, even though this is not said.

The same arises for finite group representations over a field that isn't algebraically closed ($\mathbf{Q}$, $\mathbf{R}$, $\mathbf{F}_p$...): "irreducible" versus "absolutely irreducible". And for connected semisimple algebraic groups over fields. And anywhere else rationality issues are relevant. So Lang is right (and the answer to your main question is "yes"): the word "simple" is a condition within the category under consideration (isogeny category of abelian varieties over $k$, etc.). But it's nice to add clarity by saying "$k$-simple", as Pete suggests.

If $f$ is a $k$-map between locally finite type $k$-schemes with $k$ an arbitrary field (not assumed algebraically closed) then $f$ is universally injective if and only if it is injective and $k(x)$ is purely inseparable over $k(f(x))$ for all closed points $x \in X$. The idea is that one looks for failure of injectivity on $\overline{k}$-points and unravels its consequences at the level of closed points over $k$, using that $f$ is injective.