Given a Geometric Progression (GP): (128)/(125), ..., 100, 250, 625. Find the number of terms and the first three terms.

The first term is $a = \frac{128}{125}$.
From the given terms $100, 250, 625$, we can find the common ratio $r$.

We can verify this with the next pair:

The last term is $a_n = 625$. The formula for the $n$-th term is $a_n = ar^{n-1}$.
Substitute the values of $a$, $r$, and $a_n$:

Express the numbers as powers of their prime factors: $128 = 2^7$, $125 = 5^3$, $625 = 5^4$.

For this equality to hold, the powers of the bases must match.
Equating the powers of 2:

Equating the powers of 5:

Both equations yield $n=8$.

The number of terms is $\boxed{\text{8}}$.

The first term is $a_1 = a = \frac{128}{125}$.

The second term is $a_2 = ar$.

The third term is $a_3 = ar^2$.

The first three terms are $\boxed{\frac{128}{125}, \frac{64}{25}, \frac{32}{5}}$.

The sequence is $\frac{81}{8}, \frac{27}{4}, \frac{9}{2}, ...$
The first term is $a = \frac{81}{8}$.

Verify with the next pair:

The common ratio is $r = \frac{2}{3}$.

We want to find $n$ such that $a_n = \frac{256}{2187}$.
The formula is $a_n = ar^{n-1}$.

$81 = 3^4$
$8 = 2^3$
$256 = 2^8$
$2187 = 3^7$
Substitute these into the equation:

Equating the powers of 2:

Equating the powers of 3:

Both equations give $n=12$.

The term is the $\boxed{\text{12th term}}$.