2. There is the inequality $12\sqrt{3}A\leq P$ that needs to be satisfied first of all to get a triangle. 

Using Heron's formula for a triangle with sides $x$ and $y$, you are asking for the rational solutions to $(x+y-P/2)(P/2-x)(P/2-y)=2A^2/P$. The projective closure of this curve is an elliptic curve with $O=(1:-1:0)$ and two other rational $3$-torsion points at infinity.

Now for instance the curve for $A=1$ and $P=5$ is isomorphic to the curve with Cremona label 9650m1. This curve has $E(\mathbb{Q})=\mathbb{Z}/3\mathbb{Z}$ and hence there is no triangle with rational side of perimeter $1$ and area $5$. So the answer to this question is no in general. 

For other values the rank is positive and you get solutions. For instance $P=6$ and $A=1$, we get $x=5/3$ and $y=3/2$ corresponding to the generator of the Mordell-Weil group. There are infinitely many solutions which correspond to $x>0$ and $y>0$ and $6-x-y>0$.

For $P=8$ and $A=3$, the curve has $6$ rational points, so one only gets finitely many (3) triangles in this case. $x=y=5/2$ is one.

(I hope I did not do any errors, when doing this quickly. I guess someones spots them soon.)