16. (i) Use Euler's criterion.
(ii) Use Euler's criterion and $\left(\frac{-2}{p}\right)=-1$. The solutions can be written separately as: when $a^{2 m+1} \equiv 1$ $(\bmod p)$, $x_{0}= \pm a^{m+1}$; when $a^{2 m+1} \equiv-1(\bmod p)$, $x_{0}= \pm 2^{2 m+1} a^{m+1}$.
(iii) Using the same method as (ii), the role of $b$ is equivalent to 2 in (ii). Let $p=2^{l} n+1, l \geqslant 3,2 \nmid n$. We have $a^{2^{l-1}} \equiv \pm 1(\bmod p)$ and $b^{2^{l-1} n} \equiv-1(\bmod p)$. It follows that there must be a non-negative integer $s_{1}$ such that $a^{2^{l-2} n} \cdot b^{s_{1} 2^{l-1}} \equiv 1(\bmod p)$. Therefore, $a^{2^{l-3} n} b_{1}^{s_{1} 2^{l-2}} \equiv \pm 1(\bmod p)$. Using $b^{b^{l-1} n} \equiv -1(\bmod p)$ again, we can deduce: there must be a non-negative integer $s_{2}$ such that $a^{2^{l-3} n} b^{s_{2}} 2^{l-2} \equiv 1(\bmod p)$, and $a^{2^{l-4} n} b^{t_{2}} 2^{l-3} \equiv \pm 1(\bmod p)$. This way, we finally get: there exists a non-negative integer $s_{k}$ such that $a^{n} b^{2 s_{k}} \equiv 1(\bmod p)$. Therefore, the solution is $x_{0}= \pm a^{(n+1) / 2} \cdot b^{s} k$.