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Hope this helps I'm not sure if its correct though, it must be correct anyway. So yea.

a) 25%

b) 12.5%

c) 99.9%

Step-by-step explanation:

a)

If we divide the interval [0,100] if four parts of the same length we would obtain the intervals

[0,25], [25,50], [50,75] and [75,100]

Since the offers will be uniformly distributed between 0 and 100 dollars, a given offer has 25% probability of falling in the interval [75,100] (the interval of interest)

b)

Let's call

N= “the offer is less than $75”

Y= “the offer is greater or equal than 75”

In 3 days we have 2*2*2=8 possible outcomes

(N,N,N), (N,N,Y), (N,Y,N), (Y,N,N), (N,Y,Y), (Y,N,Y), (Y,Y,N), (Y,Y,Y).  

So the probability of (N,N,Y) is 1/8 = 0.125  = 12.5%

c)

After 10 days we have possible outcomes.  

In all this offers, 1,023 are of interest and only one is not (none of the offers is $75 or more).  

So, the probability that at least one of them offers you $75 or more is

1,023/1,024 = 0.999 or 99.9%