If `alpha, beta` are the roots of the equation `x^(2)-2x-a^(2)+1=0` and `gamma, delta` are the roots of the equation
`x^(2)-2(a+1)x+a(a-1)=0` such that `alpha, beta epsilonn (gamma, delta)` find the value of `a`.

Explanation

To solve the problem, we need to find the value of \( a \) such that the roots \( \alpha, \beta \) of the first equation are also roots of the second equation.

### Step 1: Find the roots of the first equation
The first equation is given as:

\[ x^2 - 2x - (a^2 - 1) = 0 \]

We can rewrite this as:

\[ x^2 - 2x + (1 - a^2) = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -2, c = 1 - a^2 \):

\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (1 - a^2)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ b^2 - 4ac = 4 - 4(1 - a^2) = 4 - 4 + 4a^2 = 4a^2 \]

Thus, the roots are:

\[ x = \frac{2 \pm 2a}{2} = 1 \pm a \]

So, we have:

\[ \alpha = 1 + a, \quad \beta = 1 - a \]

### Step 2: Find the roots of the second equation
The second equation is given as:

\[ x^2 - 2(a + 1)x + a(a - 1) = 0 \]

Using the quadratic formula again:

\[ x = \frac{2(a + 1) \pm \sqrt{(2(a + 1))^2 - 4 \cdot 1 \cdot a(a - 1)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (2(a + 1))^2 - 4a(a - 1) = 4(a^2 + 2a + 1) - 4(a^2 - a) = 4a^2 + 8a + 4 - 4a^2 + 4a = 12a + 4 \]

Thus, the roots are:

\[ x = \frac{2(a + 1) \pm \sqrt{12a + 4}}{2} = (a + 1) \pm \sqrt{3a + 1} \]

So, we have:

\[ \gamma = (a + 1) + \sqrt{3a + 1}, \quad \delta = (a + 1) - \sqrt{3a + 1} \]

### Step 3: Establish the conditions for roots
Since \( \alpha, \beta \in \{\gamma, \delta\} \), we need to ensure:

1. \( \alpha = 1 + a \) or \( \beta = 1 - a \) must equal \( \gamma \) or \( \delta \).

We will check both cases:

#### Case 1: \( \alpha = \gamma \)
Setting \( 1 + a = (a + 1) + \sqrt{3a + 1} \):

\[ 0 = \sqrt{3a + 1} \]

This implies:

\[ 3a + 1 = 0 \implies a = -\frac{1}{3} \]

#### Case 2: \( \alpha = \delta \)
Setting \( 1 + a = (a + 1) - \sqrt{3a + 1} \):

\[ \sqrt{3a + 1} = 0 \]

This also gives:

\[ 3a + 1 = 0 \implies a = -\frac{1}{3} \]

#### Case 3: \( \beta = \gamma \)
Setting \( 1 - a = (a + 1) + \sqrt{3a + 1} \):

\[ 1 - a - (a + 1) = \sqrt{3a + 1} \]

This simplifies to:

\[ -2a = \sqrt{3a + 1} \]

Squaring both sides:

\[ 4a^2 = 3a + 1 \implies 4a^2 - 3a - 1 = 0 \]

Using the quadratic formula:

\[ a = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{3 \pm \sqrt{9 + 16}}{8} = \frac{3 \pm 5}{8} \]

This gives:

\[ a = 1 \quad \text{or} \quad a = -\frac{1}{4} \]

#### Case 4: \( \beta = \delta \)
Setting \( 1 - a = (a + 1) - \sqrt{3a + 1} \):

\[ 1 - a - (a + 1) = -\sqrt{3a + 1} \]

This simplifies to:

\[ -2a = -\sqrt{3a + 1} \implies 2a = \sqrt{3a + 1} \]

Squaring both sides:

\[ 4a^2 = 3a + 1 \implies 4a^2 - 3a - 1 = 0 \]

This gives the same solutions as before.

### Conclusion
The values of \( a \) that satisfy the conditions are:

\[ a = -\frac{1}{3}, \quad a = 1, \quad a = -\frac{1}{4} \]

However, the only common interval from all cases is:

\[ a \in \left[-\frac{1}{4}, 1\right] \]

### Final Answer
The value of \( a \) can be \( -\frac{1}{4}, -\frac{1}{3}, \text{ or } 1 \).