Given:
$a(t) = xt^{2} + yt + z $
$100a(2) = 152$ and $200a(4) = 240$
Using that, I found that $z = 1, $ $x = -0.105 $, and $y = 0.47$
The question asks for the accumulated value at time 8 of 1600 invested at time 6. So we need to find what was invested at time 0 for it to equal 1600 at time 6:
$1600= P(-0.105*6^{2} + 0.47*6 + 1) $
=> $P = 40,000$
This is already wrong... But I don't know why. I can't find A(8) without that P.
Actually everything you've done so far is right (however there's technically no need to calculate P, since the answer would simply be $1600\times\dfrac{A(8)}{A(6)}$).
With that in mind, if you analyze $A(t)$ with the coefficients you've already found, i.e. $A(t)=-0.105t^2+0.47t+1$, you'll notice there is a turning point at $t=2.238$, that $A(t)$ is decreasing for $t>2.238$, and that there is a zero at $t=6.05$. If you check, you'll get $A(6) = 0.04$, which matches up with what you found above (i.e. it gives $P=40,000$). Your conceptual error seems to be that you're assuming $A(t)$ must always be greater than 1 (which would usually be the case, but definitely isn't in this question).
So given all of the above, we can quite easily find that $A(6)=0.04$, and $A(8)=-1.96$, and therefore the answer is $1600\times\dfrac{A(8)}{A(6)}=1600\times\dfrac{-1.96}{0.04}=-\$78,400$