Taylor Series First Order Method for $x' = x-2t$ with $x(0) = 0$ given: With the step sizes h = 1, 1/2, 1/4

My solution for all step sizes came out to be zero and not sure if I am doing this right.

Since this is first order taylor series method, I'll be using: $x_{n+1}$ = $x(n) + hx'(n)$ when $x(0) = 0$.

So because x'(0) = 0, no matter what step size I use, I end up getting 0.
Your solution is correct. You can check it  by direct method too.

$x'(t)-x=-2t$;  $x(0)=0$ is Linear first order DE, whose solution is easily obtained as:

$x(t)=2t+2-2e^t$. Expanding $e^t$ by Taylor's series;

$x(t)=2t+2-2\left(1+t+\frac{t^2}{2!}+...\right)$ which on approximating up to first order term(as you are using Taylor's series method of first order or Euler's method) gives $x(t)=2t+2-2(1+t)=0$ which is the same as obtained by you.
Your solution is correct. You can check it  by direct method too.

$x'(t)-x=-2t$;  $x(0)=0$ is Linear first order DE, whose solution is easily obtained as:

$x(t)=2t+2-2e^t$. Expanding $e^t$ by Taylor's series;

$x(t)=2t+2-2\left(1+t+\frac{t^2}{2!}+...\right)$ which on approximating up to first order term(as you are using Taylor's series method of first order or Euler's method) gives $x(t)=2t+2-2(1+t)=0$ which is the same as obtained by you.