A permutation permits any element of a set to be used not at all or once. The first letter has equal chance of being a W, as it does H, I, T or E. This is a uniform distribution, so to answer your questions:

a) 60%

b) 40%

Note that these numbers should add up to 1, or 100%, because the only options are that the word begin with a consonant or a vowel.

You have no knowledge of what the 2nd and 3rd letters are, so you can simply deduce that the chance of the first letter being a vowel is or if you include y as a vowel. Likewise, the same logic applies to the numbers. The set of digits is 1, 2, 3, 4, 5, 6, 7, 8 and 9. There are four even digits and five odd digits. The probability of picking an even digit is .

Therefore, your total probability of randomly selecting a coded item with the first letter a vowel and the last digit even is the product of these two separate probabilities.

First, find the probability of the first card meeting the criteria. Greater than 2 and less than 8 implies a 3, 4, 5, 6 or 7. There are five of these in a suit and four suits in a deck making 20 possibilities out of 52. With the first card meeting that criteria, the second card's set of possibilities decreases to a possible 19 out of 51.

The probability that both cards are greater than 2 and less than 8 is the product of these two probabilities.