I did 3 Experiments and got 3 rate as the following:
Exp1: (S2O8)2-:0.05M; I-:0.05M rate:1.7x10^-5 M/s

Exp2: (S2O8)2-:0.10M; I-:0.05M rate:3.4x10^-5 M/s

Exp3:(S2O8)2-:0.05M; I-:0.10M rate:3.6x10^-5 M/s

what effect does doubling [(S208)2-] have on the rate? By what factor does the rate increase?

what effect does doubling [I-] have on the rate? By what factor does the rate increase?

How do I calculate x, y for the rate law: Rate=K[(S208)2-]^x times [I-]^y? Do I substitute the value of[(S208)2-]and [I-] from each Exp?

How do I calculate the value of the rate constant K for each Exp and the average value?

To find what happens to the rate, compare the two values you've got with the concentration of [I^-] being kept constant, that is compare the two first experiments.

When you double the concentration of S2O8^2-, what happens to the rate? Does it stay the same? Does it double? Triple?

Post what you get! :)

double the concentration of S2O8^2- -->the rate doubles.

"By what factor does the rate increase?" Is the answer increasing the concentration of S2O8^2- by a factor of 0.05 increased the rate from 1.7x10^-5 M/s to 3.4x10^-5 M/s? Does "By what factor" mean numbers or factors such as temperature, pressure, catalyst.

According to my lab manual, for Exp 1: rate=K [S2O8^2-]^x [I^-]^y, Exp 2:rate=2^x times Rate 1; Exp 3:rate=2^y times Rate 1, so I find the rate law is Rate=K [S2O8^2-]^0.5 [I^-]^0.5, but the rate law's general form is Rate=K [S2O8^2-]^x times [I^-]^y, so I find the rate law is Rate=K [S2O8^2-] [I^-]. Which one is correct to find x and y??

I find k is 6.8x10^-3 for exp1, 6.8x10^-3, 7.2x10^-3 for exp 2 and 3. Do I add up them, then divided by 3 to find the average value?

double the concentration of S2O8^2- -->the rate doubles.

"By what factor does the rate increase?" Is the answer increasing the concentration of S2O8^2- by a factor of 0.05 increased the rate from 1.7x10^-5 M/s to 3.4x10^-5 M/s?

According to my lab manual, for Exp 1: rate=K [S2O8^2-]^x [I^-]^y, Exp 2:rate=2^x times Rate 1; Exp 3:rate=2^y times Rate 1, so I find the rate law is Rate=K [S2O8^2-]^0.5 [I^-]^0.5,

I find k is 6.8x10^-3 for exp1, 6.8x10^-3, 7.2x10^-3 for exp 2 and 3. Do I add up them, then divided by 3 to find the average value?

Hm... don't rush up too much. One thing at a time.

What does double mean?
Double means 2, hence in the rate equation, you get:

R = k[S_2O_8\ ^{2-}]^x [I^-]^y

x here is equal to 2.

The order of the reaction with respect to S2O8^2- is 2, not 0.05 or 0.5.

Now, post what you get for the second part, that is how the doubling of the concentration of iodide ions affect the rate of the reaction.
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Factor here refers to the number of times. Not a catalyst, or a condition for the reaction to occur.

I'm surprised you are doing this question without knowing this... :(

You seemed to mis use the comment feature, I have deleted all of the comments, please "answer" to give follow up