Let $k$ be a finite extension of the p-adic number field $Q\_p$ and G be a connected algebraic (not affine) group over $k$. It is well-known (see e.g. [1] Proposition 3.1) that G decomposes as
$1\rightarrow G\_{aff}\cap G\_{ant}\rightarrow G\_{aff}\times G\_{ant}\rightarrow G \rightarrow 1,$ where $G\_{aff}$ is the smallest normal connected affine subgroup of $G$ such that $G/G\_{aff}$ is an abelian variety and $G\_{ant}$ is the smallest normal subgroup such that $G/G\_{ant}$ is affine. 


Then $(G\_{ant})\_{aff}$ is a connected commutative affine algebraic group, and hence admits a unique decomposition $S\times U$, where $S$ is a torus and $U$ is connected and unipotent. Let $G'$ be a quasi-complement of $S$ in $G\_{aff}/R\_u$, where $R\_u$ is the unipotent radical of $G\_{aff}$. If $\pi:G\_{aff}\rightarrow G\_{aff}/R\_u$ is the quotient map, then $\tilde{G}=\pi^{-1}(G')$ is a quasi-complement of $S$ in $G\_{aff}$. i.e. $G\_{aff}=S\tilde{G}$ with $S \cap\tilde{G}$ is finite.
Together with above exact sequence we get an isogeny (see [2] Proposition 2.2)


$\phi:(\tilde{G}\times G\_{ant})/U \rightarrow G$ such that the kernel of $\phi$ is $(\tilde{G}\cap G\_{ant})/U$. Note that $U$ is the connected component of $\tilde{G}\cap G\_{ant}$.


Q1: Is $\phi((\tilde{G}\times G\_{ant})/U)(k))=\frac{(\tilde{G}\times G\_{ant})/U)(k)}{(\tilde{G}\cap G\_{ant})/U)(k)}$ closed in $G(k)$ w.r.t. analytic topology induced from $k$.


In general $\phi$ is not surjective on $k$-rational points. We have the following exact sequence
$1 \rightarrow ((\tilde{G}\cap G\_{ant})/U)(k)\rightarrow (\tilde{G}\times G\_{ant})/U)(k) \rightarrow G(k)\rightarrow H^1(k\_s/k, (\tilde{G}\cap G\_{ant})/U)$, where $H^1$ is the first Galois cohomology.
Since $(\tilde{G}\cap G\_{ant})/U$ is a central finite affine algebraic group, $H^1$ is finite abelien group.


Q2: Does there exist a finite Galois extension $K$ of $k$ or $Q\_p$ such that $\phi$ is indeed surjective on $K$-rational points? I think this is weaker than $H^1(k\_s/k, (\tilde{G}\cap G\_{ant})/U)=0$. If it is necessary, we may change the quasi-complement $\tilde{G}$.


[1] Anti-affine algebraic groups, Michel Brion, Journal of Algebra


[2] Principal bundles, quasi-abelian varieties and structure of algebraic groups, Carlos Sancho de Salasa, , Fernando Sancho de Salas, Journal of Algebra


I am very sorry for heavy notation.



Exactness on rational points of algebraic groups
YES to Question 1. For an arbitrary homomorphism $\phi\colon G\to F$ of connected algebraic groups over $k$, not necessarily affine, where $k$ is a $p$-adic field or $k=\mathbb{R}$, the image $\phi(G(k))$ is closed in $F(k)$.


*Proof.* If $G$ is a connected $k$-group, and $X=G/H$ is a homogeneous space of $G$, then every orbit of $G(k)$ in $X(k)$ is open. Indeed, for any point $x\in X(k)$, the differential at any element $g\in G(k)$ of the map $\psi\_x\colon G(k)\to X(k)\colon\ g\mapsto g\cdot x$ is surjective (we are in characteristic 0), and we can apply the Implicit Function Theorem over $k$. Since every orbit is open, we see that every orbit is closed.


Now if $\phi\colon G\to F$ is an epimorphism of connected $k$-groups, then $F$ is a homogeneous space of $G$ with respect to the action $g\*f=\phi(g)f$, hence the orbit $\phi(G(k))$ of $1\in F(k)$ is open and closed in $F(k)$. If $\phi\colon G\to F$ is an arbitrary homomorphism of connected $k$-groups (not necessarily epimorphism), then
the image ${\rm im}\,\phi$ is Zariski closed in $F$, see Borel's book, Cor. 1.4(a), and $G\to{\rm im}\,\phi$ is an epimorphism. We see that $\phi(G(k))$ is open and closed in the closed subgroup $({\rm im}\,\phi)(k)\subset F(k)$, hence $\phi(G(k))$ is closed in $F(k)$.


NO? to Question 2. Set $H={\rm SL}\_{2,k}\times\_k {\rm SL}\_{2,k}$, $C=\{\pm 1 \}\times \{ \pm1\} $ $=Z/2Z\oplus Z/2Z$. 
Let $E$ be an elliptic curve over $k$ that has all the points of order 2 defined over $k$. We choose central embeddings $C\hookrightarrow H$ and $C\hookrightarrow E$ and set
 $G=(H\times\_k E)/C$ with respect to the diagonal embedding.
Then $G\_{\rm ant}=E$, $G\_{\rm aff}=H$, $S=1$, $\bar G=H$, $U=1$. 
In this particular case Question 2 is whether the map
$$ \phi\colon H(K)\times E(K)\to G(K) $$
is surjective for some finite extension $K/k$.
Since $H^1(K,H)=1,$ an easy cohomological argument reduces Question 2 in this case to the question whether for any elliptic curve $E$ the map
$$ E(K)\to E(K)\colon\ x\mapsto 2x $$
can become surjective after passing to some finite extension $K/k$. 


I am not an expert on elliptic curves. In a separate question you can ask to construct an elliptic curve $E$ over $k=\mathbb{Q}\_p$ such that the homomorphism 
$$ E(K)\to E(K), \quad x\mapsto 2x $$
is not surjective for any finite extension $K/k$.


