You might have to draw a mechanism to visualize the first one. You don't need to know which oxygen gets the proton to solve the question (it's the oxygen that's going to become the leaving group). What's important is what's after that, namely the glutamate residue acting as a nucleophile to kick off glucose. That means that the only thing glucose has that's new is the proton. And then a hydroxide ion derived from water nucleophilically attacks the bound galactose to free it up (you actually highlighted this phrase). Since it's this hydroxide that is derived from water, this is where the labeled O is going to be.
For the second, you can use the M1*V1 = M2*V2 formula or you can use understanding of ratios to solve the problem. Let's trace this experiment backwards using the M1*V1 = M2*V2 formula first. Okay, so in the kinetics runs, they use 1 mL of substrate solution and 1 mL of enzyme solution. So we know that the final concentration is E, the final volume is 2 mL (1 mL + 1 mL), and the initial volume was 1 mL (1 mL of the enzyme solution). So let's plug that in. M1*(1 mL) = E*(2 mL). Therefore, M1 must be 2*E. Now, that solution was made by diluting 0.1 mL of the enzyme solution into 25 mL final volume. So let's use the formula above again: M1*V1 = M2*V2. We know that we want to end up with a solution with 2*E concentration. We know that V2 is 25 mL here. And we know that V1 is 0.1 mL. So let's plug it in: M1*(0.1 mL) = 2*E*25 mL. Therefore, M1 = 500*E.
We can also use ratios to solve this. The stock solution was diluted by a factor of 250 because 0.1 mL is 1/250 of 25 mL. Then this diluted solution is further diluted by a factor of 2 to do the kinetics runs. So to reverse the whole thing, you have to concentrate it by the same factor: 2*250 = 500.