For a set to be dense in $\mathbb{R}^2$ (or in any other metric space, for that matter) it is necessary and sufficient to check that ot intersects every open disc. So, to prove that a set isn't dense, it's enough to find one open disc that includes no points of the set. For example, in (1), take $D((\frac{1}{2},0)\frac{1}{4})$ (a disk with radius $\frac{1}{4}$ around $(\frac{1}{2},0)$). It contains no point of the set in (1). Hence the set is not dense.
To prove (4), take any open disk $D((x,y),r)$. If $r all points of the disk are in the given set. Else, take $s=min\{|x|,|y|\}$ and take any point in $D((x,y),s)$. This point is both in $D((x,y),r)$ and in the given set. Hence the set is dense.
You can prove all other cases in the same manner.