(1) Not all Lagrangians can be decomposed in kinetic and potential
(2) This can be a little trickier. Sometimes you'll simply hear that if a Lagrangian does not explicity depend on time then energy is conserved but its not that simple. First of all, if [tex]\partial L/\partial t=0[/tex] then you'll have a conserved quantity. Let's see what it looks like:
[tex]
\frac{dL}{dt}=\frac{\partial L}{\partial q}\dot{q}+\frac{\partial L}{\partial \dot{q}}\ddot{q}+0[/tex]
We can replace the [tex]\partial L/\partial q[/tex] using the Euler-Lagrange equations, this gives:
[tex]
\frac{dL}{dt}=\dot{q}\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}+\frac{\partial L}{\partial \dot{q}}\ddot{q}[/tex]
Recognize the product rule and move some terms around:
[tex]
\frac{d}{dt}(\dot{q}\frac{\partial L}{\partial \dot{q}}-L)=0[/tex]
This gives us a conserved quantity which is often the energy. But not always. If you can get a hold of a copy of Classical Dyanmics by Jose and Saletan they work out the details, but the result is that the above quantity is the energy if
(1) the potential is independent of velocity
(2) the transformation from cartesian to generalized coordinates is time independent.
As for your Lagrangian, I'd be interested in knowing its motivation? Or is it just a random calculation in a textbook?