please help with these two problems, can you use excel, and explain the formulas used please. I have attached below copies of the mentioned spreadsheet.

1. Please refer to the spreadsheet titled 'Ch18_CDS_Valuation.xlsm'. Assume the following: Recovery rate =40% Notional =$1,000,000 4 quarters = Dec to March, March to June, June to September and September to December. Zero price = Use the zero prices below Survival probability Compute the CDS spread. 2. Given the above probability structure, what is the probability of a default in QTR 3 if not default has earlier? What is the estimated probability a default occurrence in QTR 3 independent of the history? T=1 : Probability of default =p Probability of survival =(1p) T=2P Probability of default in the second period but not in the first period =p(1p) Probability of survival in first period and second period =(1p)(1p)=(1p)2 Probability default in the second period but not in the first period can also be written as (1p)(1p)2=p(1p) T=3 : Probability of default in the third period but not in the first two periods =(1p)(1p)p=(1p)2p Probability of survival in first period, second and third period =(1p)(1p)(1p)= (1p)3 Probability default in the second period but not in the first period can also be written as (1p)2(1p)3=(1p)2p 1. Please refer to the spreadsheet titled 'Ch18_CDS_Valuation.xlsm'. Assume the following: Recovery rate =40% Notional =$1,000,000 4 quarters = Dec to March, March to June, June to September and September to December. Zero price = Use the zero prices below Survival probability Compute the CDS spread. 2. Given the above probability structure, what is the probability of a default in QTR 3 if not default has earlier? What is the estimated probability a default occurrence in QTR 3 independent of the history? T=1 : Probability of default =p Probability of survival =(1p) T=2P Probability of default in the second period but not in the first period =p(1p) Probability of survival in first period and second period =(1p)(1p)=(1p)2 Probability default in the second period but not in the first period can also be written as (1p)(1p)2=p(1p) T=3 : Probability of default in the third period but not in the first two periods =(1p)(1p)p=(1p)2p Probability of survival in first period, second and third period =(1p)(1p)(1p)= (1p)3 Probability default in the second period but not in the first period can also be written as (1p)2(1p)3=(1p)2p