$1000 is invested a k, compounded continuously, and grows to $1144.54, so what is our rate of interest, ...these questions are a bit out of order... lets take 2 first....

Since the compounding is continuous, the formula would take the form of Y = Pe^(rt)... note from your previous question that I answered, we had Y=P(1+b)^x (where P was a...but here we are dealing with strictly money so by convention we use P to denote present value or principal.). As the compounding increases, that is b/m and x*m such that m is the number of periods and m approaches say ... infinity (we compounded by minutes, even by seconds, ... or miliseconds... or microseconds... etc... until we reach something so small only the mind can conceive of it ... then you would have reached infinity number of periods ... follow? ) As we increase this amount of compounding, mathematically, the second equation Y= P (1+b)^x becomes Y = Pe^(rt). ... now lets solve (1).

(1) What is the interest rate ... r... , Y = Pe^(rt).... Let's solve => ln(Y/P)= rt ln(e). Given the fact that ln(e)=1, we have ln(Y/P)=rt... thereby r=ln(Y/P)/t ... ln(1144.54/1000)/3 = 0.04500093651... or 4.5% (check these numbers with a calculator, just going off of Google.com's calculator here.

(2) answered above Y=Pe^(rt) => $1144.54=1000(e^(0.045*3))

(3)After 8 years... just change 3 to 8 in the above formula and calculate Y = 1000(e^(0.045*8) = $1433.32941

(4) When does it double? I approximate 16 years using rule of 72... lets check by solving, this time lets use the derivation above and just algebraically get t = ln(Y/P)/r ... it's pretty symmetric so this part is simple, now we double so that means that Y = 2000 (as P =1000), so t = ln(2000/1000)/0.045 = 15.40327! And we are pretty close to 16 all things considered with a rough calculation ! :D