To solve the given expressions, we need to evaluate each term separately:

1. \(\left(-\frac{1}{2}\right)^0\)
2. \(\left(3 \frac{1}{2}\right)^2\)
3. \((2.5)^3\)
4. \(\left(-\frac{1}{4}\right)^4\)

Let's calculate each one:

1. \(\left(-\frac{1}{2}\right)^0\):
Any non-zero number raised to the power of 0 is 1.
\[
\left(-\frac{1}{2}\right)^0 = 1
\]

2. \(\left(3 \frac{1}{2}\right)^2\):
Convert the mixed number to an improper fraction first.
\[
3 \frac{1}{2} = \frac{7}{2}
\]
Now, square the fraction.
\[
\left(\frac{7}{2}\right)^2 = \frac{49}{4}
\]

3. \((2.5)^3\):
Calculate the cube of 2.5.
\[
(2.5)^3 = 2.5 \times 2.5 \times 2.5 = 15.625
\]

4. \(\left(-\frac{1}{4}\right)^4\):
Calculate the fourth power of \(-\frac{1}{4}\).
\[
\left(-\frac{1}{4}\right)^4 = \left(\frac{1}{4}\right)^4 = \frac{1}{256}
\]

So, the results are:
1. \(\left(-\frac{1}{2}\right)^0 = 1\)
2. \(\left(3 \frac{1}{2}\right)^2 = \frac{49}{4}\)
3. \((2.5)^3 = 15.625\)
4. \(\left(-\frac{1}{4}\right)^4 = \frac{1}{256}\)