Answer
See explanation.
Work Step by Step
a. The person is nearsighted. She can't see faraway objects.
b. A diverging lens is needed. For an object very far from the person’s eye, the corrective lens forms a virtual image at the eye’s far point.
c. Use the lens equation.
$$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$
$$\frac{1}{\infty}+\frac{1}{-75.0cm}=\frac{1}{f}$$
$$f=-75.0cm$$
$$P=\frac{1}{f}=\frac{1}{-0.750m}=-1.33D$$
