what can we know about this kind of group Let G be a finite group,H is an arbitrary proper subgroup of G,H is solvable,but G is not solvable.then what can we know about group G?
 A:  All we know is that $G$ is not solvable.
Any $G$, even if it's nonsolvable, will have many $H\subset G$ that are solvable. For example, the trivial subgroup $e$ and the cyclic subgroups $\langle g\rangle$ for any $g\in G$ are all abelian, hence solvable. 
Edit:
As DonAntonio pointed out, I misread the question.
Here are some thoughts on the question that was actually asked:
Assume that $G$ is not simple, and of finite length.
Then $G$ has a normal subgroup $H\subset G$, and, by the condition in the question, $H$ is solvable, and so all of $H$'s composition factors are abelian. Also, $G/H$ still satisfies the condition of the question: if it were solvable, then by combining composition series, $G$ would also be, and any proper subgroup is the image of a proper subgroup of $G$, and so will still be solvable. Finally, $G/H$ has a shorter length than $G$.
Therefore, if we repeat this process enough, $G/H$ will be simple, and we have this result:

Any $G$ with this property is an extension of a solvable group by a nonabelian simple group with this property.

I suspect that the converse holds as long as $G/H$ is not a subgroup of $G$.
