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 $n = 3$ is possible if given numbers are $-1$, $1$ and $2$.

b) Let $2a+1$ be the largest odd number and $2b$ be the largest even number. The number $2a+1$ of even numbers does not exceed $b+5$ since they are all at most $2b$ and at least $-8$. At the same time the number $2b$ of odd numbers does not exceed $a+6$ since they are all no more than $2a+1$ and no less than $-9$. Whence
$$\begin{cases} 2a+1 \le b+5, \\ 2b \le a+6. \end{cases} \qquad (1)$$
Summing up these equalities we get that $a+b \le 10$. Let us estimate the largest possible value of the number $n$ depending on the sum $a+b$. If $a+b = 10$ then the first inequality of system (1) is equivalent to the inequality $3a \le 14$, so the left side is less than the right one by at least $2$, the second inequality of system (1) is equivalent to $3b \le 16$ and in it the left side is less than the right side by at least one. So the total number of numbers does not exceed $(b+5-2)+(a+6-1) = 18$.

If $a+b=9$, then the first inequality in system (1) is equivalent to the inequality $3a \le 13$, so the left side is less than the right one by at least $1$. So the total number of numbers does not exceed $(b+5-1)+(a+6) = 19$. Note that $n=19$ in this case is possible with $a=4$ and $b=5$ if given all odd numbers from $-9$ to $9$ and all even numbers from $-6$ to $10$.

If $a+b \le 8$ then $n$ doesn't exceed $(b+5)+(a+6) \le 19$. Thus, in all cases $n \le 19$ and the equality is attainable.