a) Since $1+6 = 2+5 = 3+4$, every set of $5$ numbers from $1, 2, 3, 4, 5, 6$ contains $4$ distinct numbers $a, b, c, d$ so that $a + b = c + d = 7$.

b) Indeed, no $4$ numbers out of $1, 2, 3, 5, 8$ provide equal sums: if we do not choose $8$, then $5+a > b+c$, for every $a, b, c \in \{1, 2, 3\}$, and if we choose $8$, then $8+a > b+c$, for every $a, b, c \in \{1, 2, 3, 5\}$.
The number $n = 7$ is special. Indeed:
* if we do not choose $7$, $4$, or $1$, then the argument from a) applies to the sums $1+6=2+5=3+4$, $1+7=2+6=3+5$, respectively $2+7=3+6=4+5$;
* if we choose $1$, $4$, or $7$ and two of the numbers $2, 3, 5, 6$, then we get the equal sums $1+4=3+2$, $1+5=2+4$, $1+7=2+6$, $1+7=3+5$, $3+7=4+6$, or $4+7=5+6$.
Since $5$ is, obviously, special, the special numbers are $5$, $6$ and $7$.