1. **State the problem:** We need to find the area swept by a windshield wiper of length $l=1.5$ m that moves through an angle of $75^\circ$. 2. **Formula for area of a sector:** The area $A$ of a sector of a circle with radius $r$ and central angle $\theta$ (in radians) is given by: $$A = \frac{1}{2} r^2 \theta$$ 3. **Convert angle to radians:** Since the formula requires radians, convert $75^\circ$ to radians using: $$\theta = 75^\circ \times \frac{\pi}{180^\circ} = \frac{75\pi}{180} = \frac{5\pi}{12}$$ 4. **Substitute values:** $$A = \frac{1}{2} \times (1.5)^2 \times \frac{5\pi}{12}$$ 5. **Calculate intermediate steps:** $$A = \frac{1}{2} \times 2.25 \times \frac{5\pi}{12} = \frac{2.25}{2} \times \frac{5\pi}{12} = 1.125 \times \frac{5\pi}{12}$$ 6. **Simplify multiplication:** $$A = \frac{1.125 \times 5\pi}{12} = \frac{5.625\pi}{12}$$ 7. **Final simplification:** $$A = \frac{5.625}{12} \pi = 0.46875\pi$$ **Answer:** The area swept by the wiper is $\boxed{\frac{5.625}{12} \pi \text{ m}^2}$ or approximately $0.46875\pi \text{ m}^2$.