I'll give you the answer for the growing case in i and ii

Note y = Ae^(kt)

y(0) = A

y = y(0)e^(kt) = y0 e^(kt)

if y(t1) = y1

Then y1 = y0 e^(kt1)

1/kln(y1/y0) = t1

Similarly for ii y(tn) = yn

yields 1/k ln(yn/y0) = tn

For your Lake problem You know your solution can't be correct

If P(t)=100,000e^(-t/1000) then P(0) =100,00

But if the concentration is initially .1% and there are 108 m^3

P(0) = .108m^3