I saw this question in few book but no detail, hope someone can show the working. Use the Mobius inversion formula to prove

1. $\displaystyle \sum_{d\,\mid \,n}\mu(d)\tau(\displaystyle{\frac{n}{d}})=1$

2.$\displaystyle \sum_{d\,\mid \,n}\mu(d)\sigma(\displaystyle{\frac{n}{d}})=n$

Thank you for your help.