If you're going to approximate $y(2)$ using Euler and Improved Euler, you're going to need to compute up to $y_4$ since your step size is $h=.5$. Let $f(x,y) = 2x$, $x_0=0$ and hence $x_i=x_0 + ih= .5i$. Given that the initial condition is $y(0)=1$, we have by Euler's Method that
$$y_1 = 1 + .5f(0,1) 1 + .5(2(0)) = 1$$
$$y_2 = 1 + .5f(.5,1) = 1+ .5(2(.5)) = 1.5$$
$$y_3 = 1.5 + .5f(1,1.5) = 1.5 + .5(2(1)) = 2.5$$
$$y_4 = 2.5 + .5f(1.5,2.5) = 2.5 + .5(2(1.5)) = 4$$
Thus $y_4=4$ is our approximation for $y(2)$.
Likewise, using Heun's (Improved Euler's) method
$$y_n = y_{n+1} = y_n + \frac{h}{2}\left(f(x_n,y_n) + f(x_n+h,y_n+hf(x_n,y_n))\right)$$
where $f(x,y) = 2x$, $x_0=0$, $x_i=.5i$ and $y_0=1$, computing up to $y_4$ (which I leave to you to verify) leaves us with
$$\begin{aligned}y_1 &= 1.25\\ y_2 &= 2\\ y_3 &= 3.25\\ y_4 &= 5\end{aligned}$$
And thus $y_4=5$ is our approximation for y(2)$.
Finally, for Runge-Kutta, you only need to do two iterations since the step size is $h=1$. Using the same set up as in the other two iteration methods, we see that $$y_1 = 1 + \frac{1}{6}\left(2(0) + 2[2(.5)]+ 2[2(.5)] + 2(1)\right) = 2$$
$$y_2 = 2 + \frac{1}{6}\left(2(1) + 2[2(1.5)] + 2[2(1.5)] + 2(2)\right) = 5$$
So your calculations for Runge-Kutta were correct.
