From the given information, we know that the mean and standard deviare are, respectively,

Then, we need to use the z score value in order to answer part a and b.
Part a.
In this case, we need to find the z score values for 200 and 220 calories. For the first case, we have

which gives

For the second case, we have

So, we need to find the probability between z= -2.5 and z= -0.5 on the z-table. This is shown the the following picture:
Therefore, the answer for part a is: 0.30233
Part b.
In this case, we need to find the z score value for 190 calories and get the respective probability. Then, the z score is given by

So, we need to find the following probability

which is equal to

as it is shown in the following picture:
Therefore, the answer for part b is: 0.00023263