There are 3 even numbered boxes and 4 odd numbered boxes.
a) If only one box labelled with even number can be selected, then the number of ways of choosing the boxes is (3C1) * (4C4) = 3. The number of ways to arrange these is 5! since the 5 boxes can be arranged in 5P5 ways. Hence, required answer shall be 3*5! = 360.
b) If all the even numbered boxes can be selected, then you have cases where you select 1 even 4 odd, 2 even 3 odd and 3 even 2 odd. These can be chosen in (3C1*4C4) + (3C2*4C3)+(3C3*4C2) ways, which equals 21. Again arranging them is 5! because for each case you have 5P5 ways of arranging them. Hence, the answer here will be 21*5! = 2520.