Solvable Lie groups acts on spheres Is there any connected solvable Lie group acts transitively on the sphere $\mathbb S^2$? 

The answer is no. In fact, I will prove more:

Theorem. Suppose that $G\times M\to M$ is a continuous transitive action of a connected Lie group, where  $M$ is any non-aspherical manifold (universal cover is not contractible). Then $G$ cannot be solvable.

Proof. Pick $x\in M$ and let $H<G$ denote the stabilizer of $x$ in $G$; it is a closed subgroup of $G$; necessarily a Lie subgroup. Then the orbit map $g\mapsto gx$ defines a continuous bijection $G/H\to M$. This bijection is actually a homeomorphism. (This follows from the fact that $G/H$ is a manifold, but is true in greater generality of metrizable topological spaces and locally compact group actions.)   Hence, we can identify $M$ with the base of a principal $H$-bundle $G\to M$ (with fibers homeomorphic to $H$). Now, apply the long exact sequence of homotopy groups to this bundle:
$$
...\to \pi_{i}(H)\to \pi_i(G)\to \pi_i(M)\to \pi_{i-1}(H) \to ...
$$
If $G$ were a solvable Lie group, we would have $\pi_k(G)=0$ for all $k\ge 2$ (this is a corollary of Malcev's theorem, 1945). Since $H$ is also a solvable subgroup, we have $\pi_i(G)=\pi_{i-1}(H)=0$, $i\ge 3$.
Now, if $M=S^2$ as in your question, we would conclude from this that 
$\pi_3(M)=0$. But $\pi_3(S^2)\cong {\mathbb Z}$. A contradiction. 
More generally, since $M$ is not aspherical, $\pi_i(M)\ne 0$ for some $i\ge 3$, see here.  We again get a contradiction. qed
