a) The answer is yes, even under the weaker assumptions $a^b \le b^c$, $b^c \le c^d$, $c^d \le d^a$. The least number is $b$.

We need the inequality $\sqrt[n]{n} > \sqrt[n+1]{n+1}$ which holds for all $n \ge 3$.
It is equivalent to $(1+\frac{1}{n})^n < n$ and can be obtained by standard induction. The base $n=3$ is clear, and if $(1+\frac{1}{n})^n < n$ for some $n$ then
$$\left(1 + \frac{1}{n+1}\right)^{n+1} < \left(1 + \frac{1}{n}\right)^n \left(1 + \frac{1}{n+1}\right) < n \left(1 + \frac{1}{n+1}\right) < n + 1.$$
In particular $m > n \ge 3$ implies $\sqrt[n]{m} < \sqrt[n]{n}$. We use this general inequality to prove the following claim:
If $u$, $v$, $w \in \mathbb{N}$ satisfy $u^v \le v^w$ then $w \ge u$ or $w \ge v$.
Suppose on the contrary that $w < u$, $w < v$ and write $u^v \le v^w$ as $\sqrt[v]{v} \ge \sqrt[w]{u}$. Now $w < u$ implies $\sqrt[v]{v} \ge \sqrt[w]{u} > \sqrt[w]{w}$, which can hold only if $w \in \{1,2\}$. Indeed if $w \ge 3$ then $v > w \ge 3$, so the general inequality leads to the impossible $\sqrt[v]{v} < \sqrt[w]{w}$. For $w = 2$ the condition is $u^v \le v^2$. Because $v > w = 2$, this gives $v > u \ge 3$. Therefore $\sqrt[v]{v} < \sqrt[w]{u}$ by the general inequality. On the other hand $\sqrt[w]{u} < \sqrt{2}$, so that $\sqrt[v]{v} < \sqrt[w]{u} < \sqrt{u}$. However $\sqrt[v]{v} < \sqrt{u}$ contradicts $u^v \le v^2$. Finally if $w = 1$ then $u^v \le v$. On the other hand $u > 1$, hence $u^v \ge 2^v > v$ for all $v \in \mathbb{N}$. The claim is proven.
Given $a^b \le b^c$, $b^c \le c^d$, $c^d \le d^a$, we apply the claim to the triples $(a, b, c)$, $(b, c, d)$, $(c, d, a)$. Because $a$, $b$, $c$, $d$ are distinct, the conclusion is that none of $c$, $d$ and $a$ can be the least one among $a$, $b$, $c$, $d$. Therefore the least number is $b$.

b) The answer is no. Both quadruples $a = 2^8$, $b = 2$, $c = 2^4$, $d = 2^2$ and $a = 2^8$, $b = 2$, $c = 2^4$, $d = 2^9$ satisfy the condition. The greatest number in the first is $a = 2^8$; the greatest number in the second is $d = 2^9$.