Let the principal be $P$, the amount be $A$, the annual interest rate be $R \%$, and the number of years be $n$.
The formula for compound interest compounded annually is: $\(A=P\left(1+\frac{R}{100}\right)^n\)$
Part 1: Find the value of $\(\left(1+\frac{R}{100}\right)\)$
Given:

$$
\(\begin{aligned}
& P=4000 \\
& A=16000 \\
& n=8 \text { years }
\end{aligned}\)
$$


Substitute these values into the formula:

$$
\(16000=4000\left(1+\frac{R}{100}\right)^8\)
$$


Divide both sides by 4000:

$$
\(\begin{aligned}
& \frac{16000}{4000}=\left(1+\frac{R}{100}\right)^8 \\
& 4=\left(1+\frac{R}{100}\right)^8
\end{aligned}\)
$$


Let $\(X=\left(1+\frac{R}{100}\right)\)$. So, $\(X^8=4\)$.
Part 2: Calculate the number of years for the second investment
Given:
New $\(P=2000\)$
New $\(A=16000\)$
The rate is still $\(R \%\)$, so $\(\left(1+\frac{R}{100}\right)\)$ is still $X$.
Let the new number of years be $\(n^{\prime}\)$.

Substitute these values into the formula:

$$
\(\begin{aligned}
16000 & =2000\left(1+\frac{R}{100}\right)^{n^{\prime}} \\
16000 & =2000 X^{n^{\prime}}
\end{aligned}\)
$$


Divide both sides by 2000:

$$
\(\begin{aligned}
& \frac{16000}{2000}=X^{n^{\prime}} \\
& 8=X^{n^{\prime}}
\end{aligned}\)
$$


Now we have $\(X^8=4\)$ and $\(X^{n^{\prime}}=8\)$.
We can express 8 and 4 as powers of 2 :

$$
\(\begin{aligned}
& 4=2^2 \\
& 8=2^3
\end{aligned}\)
$$


So, we have:

$$
\(\begin{aligned}
& X^8=2^2 \\
& X^{n^{\prime}}=2^3
\end{aligned}\)
$$


From $\(X^8=2^2\)$, we can find $X$ :

$$
\(X=\left(2^2\right)^{1 / 8}=2^{2 / 8}=2^{1 / 4}\)
$$


Now substitute this value of $X$ into the second equation:

$$
\(\begin{aligned}
& \left(2^{1 / 4}\right)^{n^{\prime}}=2^3 \\
& 2^{n^{\prime} / 4}=2^3
\end{aligned}\)
$$


Equating the exponents:

$$
\(\begin{aligned}
\frac{n^{\prime}}{4} & =3 \\
n^{\prime} & =3 \times 4 \\
n^{\prime} & =12
\end{aligned}\)
$$


The investment will become $\(\$ 16000\)$ in 12 years.