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 \( a \) given the equations and the conditions on the roots. Let's break down the solution step by step.

### Step 1: Identify the roots of the first equation

The first equation is:

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

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 \( a = 1, b = -2, c = 1 - a^2 \):

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

Thus, the roots \( \alpha \) and \( \beta \) are:

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

### Step 2: Identify the roots of the second equation

The second equation is:

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

Using the quadratic formula again, where \( a = 1, b = -2(a + 1), c = a(a - 1) \):

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

Calculating the discriminant:

\[ (a + 1)^2 - a(a - 1) = a^2 + 2a + 1 - (a^2 - a) = 3a + 1 \]

Thus, the roots \( \gamma \) and \( \delta \) are:

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

### Step 3: Set the conditions for the roots

We are given that \( \alpha, \beta \in (\gamma, \delta) \). This means:

\[ \gamma < \alpha < \beta < \delta \]

Substituting the expressions for \( \alpha \) and \( \beta \):

1. **Condition 1:** \( \gamma < \alpha \)
\[ (a + 1) + \sqrt{3a + 1} < 1 + a \]
Simplifying gives:
\[ \sqrt{3a + 1} < 0 \quad \Rightarrow \quad 3a + 1 < 0 \quad \Rightarrow \quad a < -\frac{1}{3} \]

2. **Condition 2:** \( \beta < \delta \)
\[ 1 - a < (a + 1) - \sqrt{3a + 1} \]
Simplifying gives:
\[ \sqrt{3a + 1} < 2a \quad \Rightarrow \quad 3a + 1 < 4a^2 \quad \Rightarrow \quad 4a^2 - 3a - 1 > 0 \]

Using the quadratic formula on \( 4a^2 - 3a - 1 = 0 \):

\[ a = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} = \frac{3 \pm \sqrt{9 + 16}}{8} = \frac{3 \pm 5}{8} \]
Thus, the roots are \( a = 1 \) and \( a = -\frac{1}{4} \).

The inequality \( 4a^2 - 3a - 1 > 0 \) holds for \( a < -\frac{1}{4} \) or \( a > 1 \).

### Step 4: Combine conditions

From the conditions derived:

1. \( a < -\frac{1}{3} \)
2. \( a < -\frac{1}{4} \) or \( a > 1 \)

The intersection of these conditions gives:

\[ a < -\frac{1}{3} \]

### Conclusion

Thus, the value of \( a \) that satisfies all conditions is:

\[ \boxed{(-\infty, -\frac{1}{3})} \]