The future value of an amount P invested at interest rate r (in decimals), compounded n times per year and for n years is given by:

Here \(P=3000,\ r=8\%,\ n=4,\ t=6\)

Plugging the values,

Hence the future value is \($4825.31\)

The interest earned is

\(A=P\left(1+\frac{r}{n}\right)^{nt}\)

\(A=(3000)(1+\frac{0.08}{4}^{(4)(6)})\)

\(A=(3000)(1+0.02)^{24}\)

\(A=(3000)(1.02)^{24}\)

\(A=(3000)(1.60844)\)

\(A=4825.31\)

\(I=4825.31-3000=1825.31\)