a) Let $d$ be the greatest common divisor of some four different two-digit numbers. Since all these numbers are divisible by $d$, the least possible candidates of these four numbers are $d$, $2d$, $3d$, $4d$. Hence $4d < 100$, thus $d \le 24$. On the other hand, the greatest common divisor of $24$, $48$, $72$ and $96$ is $24$.

b) The numbers $99$, $98$, $97$, and $95$ are pairwise relatively prime, hence $\text{lcm}(99, 98, 97, 95) = 99 \cdot 98 \cdot 97 \cdot 95$. To show that this is the largest possible, consider four different two-digit numbers $a_1$, $a_2$, $a_3$, $a_4$; assume without loss of generality that $a_1 > a_2 > a_3 > a_4$. If $a_4 \le 95$, then $\text{lcm}(a_1, a_2, a_3, a_4) \le a_1 a_2 a_3 a_4 \le 99 \cdot 98 \cdot 97 \cdot 95$. If $a_4 = 96$, then the four numbers can only be $99$, $98$, $97$, $96$, but $\text{lcm}(99, 98, 97, 96) = \frac{99 \cdot 98 \cdot 97 \cdot 96}{2 \cdot 3} < 99 \cdot 98 \cdot 97 \cdot 95$.