Please give your progress on your first question. I think you will be able to get it without much help.
I can offer some help on your second question right away. For any distance function $d$, you always have the open ball $B(x,\delta)=\{y\in X\mid d(x,y)<\delta\}$ (We will always use $\delta>0$ to make it nondegenerate.) Different $d$'s may give different shapes of the ball, but we can think of them all as ball-shaped regions around $x$.
Perhaps you've seen $d_1(x,y)=\max({|x|,|y|})$ used in $\mathbb{R}^2$, which results in a square-shaped ball around a point, or $d_2(x,y)=|x|+|y|$ resulting in a diamond shaped ball around a point.
The notion of complement, of course, will never change. It is just a set theoretic definition, and doesn't depend on any topology!
The notion of "open set" associated with this metric $d$ is entirely dependent upon these ball-regions. A set $\mathcal{O}$ is open in that topology if each each point of $\mathcal{O}$ is the center of an open ball contained in $\mathcal{O}$.
Once you feel you understand the open sets, closed sets will be exactly the complements of open sets.
You can convince yourself that in a metric space, a set is dense (intersects every nonempty open set nontrivially) iff it intersects every ball (with positive radius of course) nontrivially. If you get this, I think you should already understand how this works for any metric.
The same goes for perfectness: a set $Y$ will be perfect iff $B(x,\delta)\setminus\{x\}\cap Y\neq\emptyset$ for every $x\in Y$.
Again, the shape of the ball may change with the metric, but you use the ball uniformly in all of these concepts.