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?
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: $$ \lim_{n \to \infty} Prob\{(\sim P_1^n) \cup (\sim P_2^n)\} \le \lim_{n \to \infty} Prob\{\sim P_1^n\} + Prob\{\sim P_2^n\} \\ = \lim_{n \to \infty} Prob\{\sim P_1^n\} + \lim_{n \to \infty} Prob\{\sim P_2^n\} \\ = 0 + 0 \\ = 0. $$