Answer
See below.
Work Step by Step
a) Plugging in $n=1$ we get: $A_1=2000(1+\frac{0.024}{12})^1\approx2004$.
Plugging in $n=2$ we get: $A_2=2000(1+\frac{0.024}{12})^2\approx2008.01$.
Plugging in $n=3$ we get: $A_3=2000(1+\frac{0.024}{12})^3\approx2012.02$.
Plugging in $n=4$ we get: $A_4=2000(1+\frac{0.024}{12})^4\approx2016.05$.
Plugging in $n=5$ we get: $A_5=2000(1+\frac{0.024}{12})^5\approx2020.08$.
Plugging in $n=6$ we get: $A_6=2000(1+\frac{0.024}{12})^6\approx2024.12$.
b) Three years is the same as $3\cdot12=36$ months, so plugging in $n=36$ we get:
$A_{36}=2000(1+\frac{0.024}{12})^{36}\approx2149.16$.
