Relativistic mass and energy Why can't we use $\frac{1}{2}(\gamma m)v^2$ to calculate kinetic energy of a particle moving at Relativistic speed?
 A: Actually, "relativistic mass" is a historic name, but it is no longer used today (at least it shouldn't). 
The idea of relativistic mass was create because special relativity shows that inertia of a given body increases with its velocity.
At low speeds we still have $\text{inertia}=\text{mass}$, but this is no longer true at high speeds.
In fact, classicaly we have
$$\vec{p}=m\vec{v}$$
But in special relativity it turns
$$\vec{p}=\gamma\ m \vec{v} \qquad \therefore \qquad \gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}$$
So, for didatical reasons, was  taught that we have two types of mass:


*

*rest mass $m_0$: mass measured in a frame at rest in relation to the frame's particle.

*relativistic mass $m=\gamma\ m_0$.


Even Feynmann used this notation in your lectures, because this idea is very intuitive. But you have seen that we cannot use this "relativistic correction" all the time, because sometimes it takes us the wrong way. (kinetic energy example)
In fact, today we are stopping using it, to avoid this kind of mistake. You could see this in more details here (video by Fermilab channel)
A: Yes, because relativity isn't encapsulated in replacing every $m$ in your formulas with the "relativistic mass" $\gamma m$. 
Relativity isn't a small correction that you can post-facto apply in all Newtonian formulas by uniformly changing some parameters. Relativistic effects, while simple, are subtle and canonical. You have to start from the basic principle, i.e., Lorentz invariance of the physical laws, and formulate a relativistic version of the whole of mechanics. See, for example, "Gravitation and Cosmology", Weinberg, S., Chapter $2$. 

According to the principles of relativity (i.e., the requirement that the laws of nature be invariant under Lorentz transformations), the energy and the momentum are given by
$$\begin{align}
E&=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}&=\gamma mc^2\tag{1}\\
p&=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}&=\gamma mv\tag{2}\\
\end{align}$$
These are the energy and momentum because these are the conserved quantities obtained from the Lorentz invariant action $S=-\int d\tau\,mc^2$ where $\tau$ is the proper time of the particle. 
In a tragic turn of events, a misconception of relativistic mass was introduced to sell bumper-stickers with $E=Mc^2$ written on them. You see, you can define a relativistic mass $M=\gamma m$ and "simplify" the expressions of energy and momentum to read $E=Mc^2$ and $p=Mv$ (notice that the trick of replacing mass with the relativistic mass in the Newtonian formula would only work for the expression for momentum, not energy--hence, your confusion). This looks simple but is, in fact, horrible. You have introduced a completely unnatural quantity out of your own will. So, what happens? You run into trouble. For example, in order to treat the dynamics in this spirit, you have to define two kinds of relativistic mass: one for when the force acts along the direction of motion and one for when the force acts perpendicular to the motion. And if the force is acting at an angle that is neither perpendicular nor parallel to the direction of motion, you have to do the calculation in two components and use the longitudinal relativistic mass for one calculation and the transverse relativistic mass for the other, simultaneously. This makes it clear that this is no way to go forward. Equations $(1)$ and $(2)$ are nice enough and well-motivated from physical principles. Stick to them!  
A: The formula $E=\gamma mc^2$ applies to particles moving at any speed.
At $v=0$ , $\gamma$ is 1. So we get the famous $E=mc^2$. 
At low velocities, we can approximate $\gamma$ as $1 + \frac12\frac{v^2}{c^2}$. Substituting this approximation we get $E=mc^2 + \frac12 mv^2$ .
For large velocities $\gamma$ approaches infinity, and this is why massive particles can't travel at light speed.
