a) The largest odd and even numbers are positive integers since there are numbers of both parities. This means that the largest even number is not less than two, the largest even number is not less than one, and the total number of numbers is not less than three. Note that $n = 3$ could be, for example, if $a = 3$ and the numbers $-1$, $1$ and $2$ are given.

b) Let $2x + 1$ be the largest odd number and $2y$ be the largest even number. The number $2x + 1$ of even numbers does not exceed $y + \lfloor \frac{a+1}{2} \rfloor$, since they are all at most $2y$ and at least $-a + 1$. And the number $2y$ of odd numbers does not exceed $x + 1 + \lfloor \frac{a}{2} \rfloor$, because they are not greater than $2x + 1$ and not less than $-a + 1$. Hence
$$\begin{cases} 2x + 1 \le y + \lfloor \frac{a+1}{2} \rfloor, \\ 2y \le x + 1 + \lfloor \frac{a}{2} \rfloor. \end{cases} \quad (1)$$
Summing up these equalities we get $x + y \le \lfloor \frac{a+1}{2} \rfloor + \lfloor \frac{a}{2} \rfloor = a$, therefore $n = 2x + 1 + 2y \le 2a + 1$. If $n = 2a + 1$ then the inequalities in (1) must be the equalities and in this case
$$\begin{cases} 3x = (x + y) + \lfloor \frac{a-1}{2} \rfloor, \\ 3y = (x + y) + \lfloor \frac{a+2}{2} \rfloor. \end{cases}$$
However this is impossible, since the right-hand sides of these equalities give different remainders when divided by $3$.

Since the number $n$ is odd and $n < 2a + 1$, then $n \le 2a - 1$. Let us show that this estimate is attainable. For even $a$ one can take $x = a/2 - 1$ and $y = a/2$, and for odd $a$ one can take $x = y = (a - 1)/2$, then the system (1) will be satisfied and, obviously, there will be corresponding $n$-tuples of numbers satisfying the condition.