If almost all random graphs $G \in g(n,p)$ have property $P_1$ and $P_2$ then almost all graphs have property $P_1 \cap P_2$ 
If almost all graphs $G \in g(n,p)$ have a graph property $P_1$ and
  almost all graphs $G \in g(n,p)$ have a graph property $P_2$ then
  almost all $G \in g(n,p)$ have the property $P_1 \cap P_2$.

Where $g(n,p)$ is a Erdos-Renyi random graph. I know that we say that almost all graphs have a property $P$ in a random graph if the probability of a graph having that probability tends to 1 when the number of nodes goes to infinity.
I think I should  use the complementary properties some how but I'm not sure how to prove this? 
 A: I'd recommend looking at the complement. If you can show that the probability of the set not having property $P_1 \cap P_2$ goes to zero, you'll be done. That set consists of $(\sim P_1) \cup (\sim P_2)$, where I've used $\sim$ to denote negation. 
Now you already know that both $\sim P_1$ and $\sim P_2$ have probability that tends to zero as the number of nodes grows, because the probabilities of their complements go to one. Let's give those things names. Let's say that $P_1^n$ denotes the probability that $P_1$ holds on graphs of $n$ nodes, OK? 
And in general, we know that
$$
Prob\{A \cup B\} = Prob\{A\} + Prob\{B\} - Prob\{A \cap B\} \le Prob\{A\} + Prob\{B\}
$$
so you can conclude that 
$$
Prob\{(\sim P_1^n) \cup (\sim P_2^n)\} \le Prob\{\sim P_1^n\} + Prob\{\sim P_2^n\} 
$$
Let's look at that:
$$
\lim_{n \to \infty} Prob\{(\sim P_1^n) \cup (\sim P_2^n)\} \le 
\lim_{n \to \infty}  Prob\{\sim P_1^n\} + Prob\{\sim P_2^n\} \\
=  \lim_{n \to \infty}  Prob\{\sim P_1^n\} + \lim_{n \to \infty}  Prob\{\sim P_2^n\} \\
= 0 + 0 \\
= 0. 
$$
