Suppose G is an algebraic group defined over F, the algebraic closure of F is K. Consider the Zariski topology on G(K), is G(F) Zariski dense in G(K)?

If $G$ is connected and reductive over an arbitrary field $k$, then $G$ is $k$unirational. Thus if $k$ is moreover infinite, $G(k)$ is Zariskidense in $G(\overline{k})$. This is due to Rosenlicht (Ann. Mat. Pura Appl. (4) 43 (1957), 2550; MR0090101) in the case of perfect $k$ and Grothendieck (SGAIII) in the arbitrary case. Certainly one wants to restrict to linear groups, since an abelian variety over a global field can have a finite number of rational points. Perhaps someone else can speak to the linear, nonreductive case. 


To amplify Pete's answer, there is a reasonable discussion in Section 18 of the second edition of Borel's Linear Algebraic Groups (Springer GTM 126). In particular, his Corollary 18.3 following a discussion of unirationality in linear algebraic groups defined over a field $k$ states: Let $G$ be connected, $k$ infinite. If either $k$ is perfect, or $G$ is reductive, $G(k)$ is Zariskidense in $G$. He goes on to note an example given by Rosenlicht in the paper cited by Pete, giving a onedimensional unipotent group over an infinite but imperfect field $k$ for which $G(k)$ fails to be Zariskidense. I'm not sure what further results are out there in the literature, but one certainly has to be cautious. 


A (smooth, connected) unipotent group $U$ is said to be $k$split if there is a filtration by $k$subgroups for which the successive quotients are isomorphic to $\mathbf{G}_{a/k}$. The examples mentioned in comments (e.g. the subgroup of $\mathbf{G}_a^2$ defined by $y^p  y = tx^p$) are nonsplit unipotent groups. Any $k$split unipotent group $U$ is even a rational variety (in fact, $k$isomorphic as a variety to $\mathbf{A}^n$) so it is clear that $U(k)$ is Zariski dense in $U(k_{alg})$ when $k$ is infinite. More generally, let $G$ be a (smooth) linear algebraic group over $k$ and assume that the unipotent radical of $G$ is defined and split over $k$ (both of these conditions can fail). Then as a $k$variety, $G$ is just the product of its reductive quotient $G_{red}$ and its unipotent radical (result of Rosenlicht). In particular, $G$ is unirational and if $k$ is infinite, $G(k)$ is dense in $G(k_{alg})$. Of course, this observation isn't that interesting  in some sense, it just "identifies" the problem. 

