A straight line passes through the point of intersection $x-2y-2=0$   and  $2x-by-6=0$ and the origin. Then the complete set of values of b for which the acute angle between this line and $y=0$  is less than ${45}^0$  is

Line passes through the point of intersection of  $x-2y-2=0$ and  $2x-by-6=0$
So, its equation is given by  $\lambda (x-2y-2)+(2x-by-6)=0$
As it passes through the origin
$-2\lambda -6=0\Rightarrow \lambda =-3$
$\therefore$ Equation of the line is $-x+(6-b)y=0$ 
Its slope is  $\frac{1}{6-b}$
As its angle with $y=0$  is less than  $\frac{\pi }{4}$
$\therefore -1<\frac{1}{6-b}<1$ 
 $\Rightarrow 6-b>1or<-1$
 $\Rightarrow b<5orb>7$
But $b\ne 4$ (as the lines intersect)
 $\therefore b\in (-\infty ,4)\cup (4,5)\cup (7,\infty )$