If Lagrangian is separable in generalized coordinates and velocity does that mean Hamiltonian is equal to total energy? I know Hamiltonian is equal to total energy if Lagrangian is time independent and the potential does not depend on velocities, but does that mean Lagrangian is separable in generalized coordinates and velocities?
 A: I am not sure to understand  what you mean for separable, however your initial claim 

Hamiltonian is equal to total energy if Lagrangian is time independent and the potential does not depend on velocities

is false in general. Furthermore the explicit expression of the kinetic energy is not separable into sum of products of generalized coordinates and their time derivatives, but it may have also an added part which is function of $q$ but not of $\dot{q}$.
(Notice that a similar part already exists in $L$ and is given by the potential energy, here instead it is also included in the kinetic energy).
Consider, as the simplest example, a particle $P$ of mass $m>0$ constrained to move along the $x'$-axis (without friction) and connected to the origin by an ideal spring   of constant $k>0$. Denote by $s$ the coordinate of  $P$ along $x'$.
Finally assume that the frame $K'$ defined by $x'y'z'$ is rotating around $z$ with constant angular velocity $\Omega>0$ with respect to an inertial reference frame $K$ with axes $xyz$ and $z=z'$.
It is easy to prove that the velocity of $P$ in $K$ is
$${\bf v}|_K = s \Omega\left(-\sin (\Omega t) {\bf e}_x + \cos (\Omega t) {\bf e}_y\right)+ \dot{s}(\cos(\Omega t) {\bf e}_x + \sin (\Omega t) {\bf e}_y)$$
so that the kinetic energy in $K$ is, with trivial computations,
$$T|_K = \frac{m}{2}{\bf v}|_K^2= \frac{m}{2}\left(\dot{s}^2 + s^2\Omega^2\right)\:.$$
You already see that here $T|_K$ is not separable in $s$ and $\dot{s}$ in the sense you meant.
The Lagrangian is therefore
$$L|_K(s, \dot{s}) =  \frac{m}{2}\left(\dot{s}^2 + s^2\Omega^2\right) - \frac{k}{2}s^2\:.$$
Even if $L|_K$ is not an explicit function of time, the Hamiltonian function (which is constant in time along the solutions of Euler-Lagrange equations) does not coincide with the total energy in $K$  however!
In fact it is 
$$H(s, \dot{s}) = \frac{\partial L|_K}{\partial \dot{s}}\dot{s}- L|_K =  \frac{m}{2}\dot{s}^2 - \frac{m}{2}s^2\Omega^2\ + \frac{k}{2}s^2$$
whereas the energy in $K$ is
$$E|_K(s, \dot{s}) = T|_K + U|_K =  \frac{m}{2}\dot{s}^2 + \frac{m}{2}s^2\Omega^2\ + \frac{k}{2}s^2\:.$$
The latter is not constant along the solutions of the equation of motion.
The fundamental added hypothesis you omitted to obtain both your claimed results is that
the position of the points of the system in the rest frame used to compute velocities is a function only of generalized coordinates and not of time. 
In this case the hypothesis is violated because the position of $P$ in $K$ reads
$${\bf x}(t,q) = s\cos(\Omega t){\bf e}_x + s\sin(\Omega t){\bf e}_y$$
where $t$ explicitly appears.
Regarding the meaning of $H$ in the considered example, it is nothing but 
the total energy computed in $K'$. 
In $K'$, in addition to the force of the spring,  two further inertial forces show up: Coriolis' force that plays no role because exactly as the reactive force is normal to the velocity of the point in $K$, and the centrifugal force that acts as a repulsive spring of constant $-m\Omega^2$. As a matter of fact:
$$L|_{K'}(s, \dot{s}) =  \frac{m}{2}\dot{s}^2  - \frac{k}{2}s^2 + s^2\Omega^2 = L|_{K}(s, \dot{s})\:.$$ and thus
$$H(s, \dot{s}) = \frac{\partial L|_{K'}}{\partial \dot{s}}\dot{s}- L|_{K'} =  \frac{m}{2}\dot{s}^2  +\left( \frac{k}{2}s^2 - \frac{m}{2}s^2\Omega^2\right) = T|_{K'}+ U|_{K'} = E|_{K'}(s, \dot{s}).$$
Notice that the kinetic energy is just
$$T|_{K'}=\frac{m}{2}\dot{s}^2 $$
and thus it is separable as you claimed. In fact the position of $P$ in $K'$ is trivially $t$-indepedent
$${\bf x}(t,s) = s {\bf e}_{x'}\:.$$
The theorem you are looking for reads
THEOREM (sometimes  known as Jacobi's theorem) If, for as system admitting Largangian description, the Lagrangian does not explicitly depend on time, then the Hamiltonian function is constant in time along the solutions of E.L.
Furthermore, if in the reference frame $K$ used to construct the Lagrangian
(a) all forces (barring reactive ones)  admit potential energy independent form time and
(b) the positions of the points of the systems in $K$ are not explicit function of time,
then the following facts hold. 
(1) The Hamiltonian function coincides to the total energy.
(2) Kinetic energy is separable in the sense that it has the form (where $q =(q^1,\ldots, q^n)$)
$$T|_K(q, \dot{q}) = \sum_{h,k=1}^n a_{hk}(q) \dot{q}^h \dot{q}^k$$
so that the Lagrangian has the form
$$L|_k(q, \dot{q}) = \sum_{h,k=1}^n a_{hk}(q) \dot{q}^h \dot{q}^k - U(q)\:,$$
where $$a_{hk}(q) = \sum_{i=1}^N\frac{m_i}{2} \frac{\partial {\bf x}_i}{\partial q^h} \cdot \frac{\partial {\bf x}_i}{\partial q^k}$$
where $N$ is the number of particles of the system with masses $m_i$ and position vector ${\bf x}_i={\bf x}_i(q^1,\ldots, q^n)$ in $K$, and $n$ the number of degrees of freedom of the system.
(3) Total energy is constant along the solutions of E-L equations.
