a) For $a < b$, $\log_a(\log_a b) = (\log_a b)(\log_b(\log_a b)) > \log_b(\log_a b)$ because $\log_b(\log_a b) > 0$ and $\log_a b > 1$.
For $a > b$, the claim is reached from $\log_a b < 1$ and $\log_b(\log_a b) < 0$.

b) Induct on $n$. For $n = 2$,
$$\log_{a_1}(\log_{a_1}(a_2) + \log_{a_2}(\log_{a_2}(a_1))) > \log_{a_2}(\log_{a_1}(a_2) + \log_{a_1}(\log_{a_1}(a_1))) = \log_{a_2}(1) = 0.$$
Assume now that the claim holds for some $n$ and consider the numbers $a_1 > a_2 > \dots > a_{n+1} > 1$. Then
$$\begin{align*} & \log_{a_1}(\log_{a_1}(a_2) + \dots + \log_{a_n}(\log_{a_n}(a_{n+1}) + \log_{a_{n+1}}(\log_{a_{n+1}}(a_1))) = \\ & \qquad \log_{a_1}(\log_{a_1}(a_2) + \dots + \log_{a_{n-1}}(\log_{a_{n-1}}(a_n) + \log_{a_n}(\log_{a_n}(a_1))) + \\ & \qquad \phantom{\log_{a_1}(\log_{a_1}(a_2) + \dots + \log_{a_{n-1}}(\log_{a_{n-1}}(a_n) + \log_{a_n}(\log_{a_n}(a_1)) =} \\ & > \log_{a_{n+1}}(\log_{a_n}(a_{n+1}) + \log_{a_{n+1}}(\log_{a_{n+1}}(a_1)) - \log_{a_n}(\log_{a_n}(a_1)) > \\ & \qquad \phantom{> \log_{a_{n+1}}(\log_{a_n}(a_{n+1}) + \log_{a_{n+1}}(\log_{a_{n+1}}(a_1)) - \log_{a_n}(\log_{a_n}(a_1)) =} \\ & = \log_{a_{n+1}}(\log_{a_n}(a_1) - \log_{a_n}(\log_{a_n}(a_1))) > 0, \end{align*}$$
because $\log_{a_n} a_1 > 1$ and $a_{n+1} < a_n$.