I'm not sure from the comments above if you've solved this or not already, but I had this idea when I first read your post.
I think you can partition the outcome space, based on when $B,C,D$ die. There are four disjoint cases:
$B_0:$ none of them dies before 90,
$B_1:$ exactly one of them dies before 90,
$B_2:$ exactly two...
$B_3:$ exactly three...
So $$P(B_n)=\binom{3}{n}(1/3)^n(2/3)^{3-n}.$$
Then let $F$ be the event that $A$ dies first, and $E$ be the event that it happens before $90$.
So $$P(F\cap E)=\sum_{n=0}^3 P((F\cap E)|B_n)P(B_n)$$
So for example, $P((F\cap E)|B_0)$ is the probability that $A$ dies first and before $90$, given that the other three die after $90$. This occurs with probability $(1/3)$, since all the others are given to die after $90$. Then $P((F\cap E)|B_1)$ is the probability that $A$ dies first and before $90$, when exactly one other person dies before $90$. This probability is $(1/3)(1/2)$. The other cases are similar, and then one can just add them up.