Different approaches work for 1) and 2).
$S$ is a subspace iff whenever $x,y \in S$ and $\lambda \in \mathbb{R}$, then (1) $x+y \in S$, and (2) $\lambda x \in S$.
For $U$, following @babgen, notice that $x=(0,0,1,0) \in U$, but $(-1)x = (0,0,-1,0)$ is not. So it fails (2), hence it is not a subspace.
For $V$, you can work from the definition. For example, choose $(a,b,c,d), (a',b',c',d') \in V$. Since they are in $V$, you know that $c+2d=0$ and $a+b-2c=0$. Similarly, you know that $c'+2d'=0$ and $a'+b'-2c'=0$. Adding together you see that $(c+c')+2(d+d') =0$ and similarly for the other constraint. So it follows that $(a+a',b+b',c+c',d+d') \in V$. So Property (1) of the definition is true. You can prove the other property similarly.