The values of a for which one root of the equation $x^2-(a+1)x+a^2+a-8$ $=0$ exceeds 2 and the other is lesser than 2, are given by

given equation is $x^2-(a+1)x+a^2+a-8$ $=0$ …. ( 1 )

If the roots of the given equation are 

real and distinct D$>$o$\Rightarrow$$3a^2+2a-33<0$ $\Rightarrow$a$\in$($\frac{-11}{3}$,3)………(1)

 and   f(2)  less than 0  

$\Rightarrow 4-\left(a+1\right)2+a^2+a-8<0$

so  a$\in$(-2,3)…….       ……..(2)

from 1&2 we get   a$\in$(-2,3)