First, let's sketch the problem with the forces acting on the book:
1)
Since the book remains motionless, the sum of forces is equal to zero, in the 15° direction (table surface) and in the direction perpendicular to the table.
In the table direction, we have the frictional force and a component of the weight force:
[tex]\begin{gathered} W\cdot\sin (\theta)-F_f=0 \\ m\cdot g\cdot\sin (15\degree)-F_f=0 \\ 0.75\cdot9.8\cdot0.2588=F_f \\ F_f=1.9\text{ N} \end{gathered}[/tex]The frictional force is equal to 1.9 N.
2)
In the perpendicular direction, we have:
[tex]\begin{gathered} N-W\cdot\cos (\theta)=0 \\ N=W\cdot\cos (15\degree) \\ N=m\cdot g\cdot\cos (15\degree) \\ N=0.75\cdot9.8\cdot0.9659 \\ N=7.1\text{ N} \end{gathered}[/tex]The normal force is equal to 7.1 N.