The first step is to write the final amount in terms of the interest rate. This is a geometric series, so we can use the corresponding formulae. It's not exactly clear what the situation is, but if you're depositing at the end of each month, and getting your money back at the end of the 60th month (in which case the last payment is immediately returned), you have
sum k goes from 0 to 59 of (1000*r**k)
If you have to wait a month after the last payment to get your money back, then it would be
sum k goes from 0 to 59 of r*(1000*r**k)
If you pay at the beginning of each month, and get the money back at the end of the 60th month, then it's
sum k goes from 1 to 60 of (1000*r**k)
(Note: you used r for the interest rate. I believe that usually i is used for interest, and r is used for the quantity 1+i, which is how I am using it. I'm also using it to refer to the monthly rate.)
In Excel, you could fill 0 cells with that formula, take the sum, and then use goal seek to find r such that the sum is 75000. In R, the formula is
T = 1000*sum(r**(0:59))
You can define a function that returns how far you are from your goal of 75000:
T <- function(r){1000*sum(r**(0:59))-75000}
Then you can use a root-finding function, for instance uniroot. To use uniroot, you need a guess as to something you think is less than the desired number, and a guess that's above. Clearly, you're getting some interest, so r is above 1, and you're not doubling your number, so it looks like it's below 2. So I'll use those as lower and upper bounds.
monthly_r = uniroot(T, lower = 1, upper = 2)
This gives 1.007327. This, again, is the monthly rate. If you raise that to the twelfth power, you get 1.09155. This gives 9.155% as the APY. Expressed as APR compounded monthly, it's 8.792%. As force of interest, it's 8.760%.
This can be put in closed form as 75000 = 1000*(1-r^60)/(1-r). This gives r^60-75*r+74000 = 0. You can use a polynomial solver to solve for r.