If $\alpha ,\beta$ be the roots of $x^2-2x-a+1=0$ and $\gamma ,\delta$ the roots of $x^2-2(a+1)x+a(a-1)=0$ such that $\alpha$ and $\beta$ lie in $(\gamma ,\delta )$,then the value of a.

Given equation is $x^2-2x-a+1=0$

$\begin{array}{l}\Rightarrow \left(x^2-2x+1\right)=a^2\\ \Rightarrow (x-1{)}^2=a^2\\ \Rightarrow (x-1)=\pm a\\ \Rightarrow x=1\pm \mathrm{a}\\ \Rightarrow \alpha =1+a,\beta =1-a\end{array}$

Let $f\left(x\right)=x^2-2(a+1)x+a(a-1)$

Since $\alpha$ and $\beta$ lie in $(\gamma ,\delta )$, so

$f(1+a)<0,f(1-a)<0$ 

When $f(1+a)<0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1+a{)}^2-2(1+a\left)\right(1+a)+a(a-1)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a(a-1)-(1+a{)}^2<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(a^2-1\right)-\left(a^2+2a+1\right)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-2a-2<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+1>0 ... \left(i\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a>-1\end{array}$

Also, $f(1-a)<0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1-a{)}^2-2(1+a\left)\right(1-a)+a(a-1)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1-a)\left(\right(1-a)-2(1+a)-a)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1-a)(1-a-2-2a-a)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1-a)(-1-4a)<0\\ \begin{array}{l}\Rightarrow (a-1)(4a+1)<0\\ \Rightarrow -\frac{1}{4}<a<1 ... \left(ii\right)\end{array}\end{array}$

From Relations (i) and (ii), we get

$-\frac{1}{4}<a<1$