a) There are 3 different cases for this. For when there are 3 numbers, 4 numbers, 5 numbers
There are 3 different letters with 26 to choose from. Hence $\displaystyle\binom{26}{3}$ but this doesn't take into account order so we multiply by $3!$. Then for case 3, 4, 5. Since there are 10 numbers to choose from. We multiply $\displaystyle\binom{26}{3}\times 3!$ by $10^3$, $10^4$, $10^5$ respectively for cases when we use 3, 4, 5 digits. I assumed here that the numbers can be repeated.
b) Since the character string is "B4" (a letter then a number) then B is in the 3rd position, and 4 is in the 4th position.
$_,_,B,4,_,_$ is our format. With permutations $25\times24\times1\times1\times10\times10$
Since B and 4 are fixed we have 1 for the 3rd and 5th position. And $25, 24$ since we alread chose $B$ for the 3rd character.