A possible set of values of ' $a$ ' in which for every value of  ' $a$ ' one root of the equation $x^2-(a+1)x+a^2+a-8=0$ exceeds 2 and other is less than 2, is given by

As the roots are real and distinct

$$\begin{gathered} {(a+1)}^2-4(a^2+a-8)>0 \\ 3a^2+2a-33<0 \\ (3a+11)(a-3)<0\Rightarrow -\frac{11}{3}<a<3\left(1\right) \end{gathered}$$

Also, for $2a^2-2(a+1)+a^2+a-8<0$

$a^2-a-6<0\Rightarrow (a-3)(a+2)<0$

$-2<a<3\left(2\right)$

From (1) and (2), we get  $-2<a<3\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}.}$