First, picking 3 marbles "at the same time" is the same as picking three marbles, one at a time, without replacement. Since there are 5 red and 3 white marbles, the probability the first marble chosen is red is 5/8. There are then 4 red and 3 white marbles left. The probability the second marble chosen is red is 4/7. There are then 3 red and 3 white marbles left. The probability the third marble chosen is red is 3/6= 1/2. The probability all three marbles are red is (5/8)(4/7)(1/2)= 5/28.
The probability of "at least one white" is 1 minus the probability of "all red" so 1- 5/28= 23/28.
For problem 2, if I am reading it correctly, we have 5 red marbles and n black marbles for a total of n+ 5 marbles. Now we need to consider two possible scenarios:
1) The first marble chosen is red. The probability of that is 5/(n+5). In this case, there are now 4 red and n black marbles for a total of n+ 4. In this scenario, the probability the second marble chosen is red is 4/(n+4).
2) The first marble chosen is black. The probability of that is n/(n+5). In this case, there are now 5 red and n- 1 black marbles so still a total of n+ 4. In this scenario, the probability the second marble is chosen is red is 5/(n+ 4).
Putting those together the probability that the second marble chosen is red is [5/(n+5)][4/(n+4)]+ [n/(n+ 5)][5/(n+ 4)]= 20/(n+5)(n+4)+ 5n/(n+5)(n+ 4)= (5n+ 20)/(n+5)(n+4)= 5(n+ 4)/(n+5)(n+4)= 5/(n+ 4).