To solve the equation \((X + \frac{1}{3})^4 = \frac{1}{64}\), we can start by taking the fourth root of both sides to simplify the equation.

Given:
\[
(X + \frac{1}{3})^4 = \frac{1}{64}
\]

Take the fourth root of both sides:
\[
\sqrt[4]{(X + \frac{1}{3})^4} = \sqrt[4]{\frac{1}{64}}
\]

This simplifies to:
\[
X + \frac{1}{3} = \pm \frac{1}{2}
\]

Now, we have two separate equations to solve for \(X\):

1. \(X + \frac{1}{3} = \frac{1}{2}\)
2. \(X + \frac{1}{3} = -\frac{1}{2}\)

### Solving the first equation:
\[
X + \frac{1}{3} = \frac{1}{2}
\]

Subtract \(\frac{1}{3}\) from both sides:
\[
X = \frac{1}{2} - \frac{1}{3}
\]

Find a common denominator (which is 6):
\[
X = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
\]

### Solving the second equation:
\[
X + \frac{1}{3} = -\frac{1}{2}
\]

Subtract \(\frac{1}{3}\) from both sides:
\[
X = -\frac{1}{2} - \frac{1}{3}
\]

Find a common denominator (which is 6):
\[
X = -\frac{3}{6} - \frac{2}{6} = -\frac{5}{6}
\]

### Solutions:
The solutions to the equation \((X + \frac{1}{3})^4 = \frac{1}{64}\) are:
\[
X = \frac{1}{6} \quad \text{and} \quad X = -\frac{5}{6}
\]