1) For all $n \in \mathbb{Z}^{+}$, we have $n \sqrt{17} \notin \mathbb{Z}$ then $[n \sqrt{17}] < n \sqrt{17}$ or
$$[n \sqrt{17}]^{2} < \left(n \sqrt{17}\right)^{2} \forall n.$$
This implies that
$$\begin{aligned} & 17 n^{2} - [n \sqrt{17}]^{2} \geq 1 \\ & \Leftrightarrow 17 n^{2} - (n \sqrt{17} - \{n \sqrt{17}\})^{2} \geq 1 \\ & \Leftrightarrow 17 n^{2} - \left(17 n^{2} - 2 n \sqrt{17} \{n \sqrt{17}\} + \{n \sqrt{17}\}^{2}\right) \geq 1. \\ & \Leftrightarrow \{n \sqrt{17}\} \geq \frac{1 + \{n \sqrt{17}\}^{2}}{2 n \sqrt{17}} > \frac{1}{2 n \sqrt{17}} \end{aligned}$$

2) Consider the Pell equation $m^{2} - 17 n^{2} = -1$, since $17$ is the prime of form $4k+1$ then this equation has infinitely many positive integer solutions.
Thus $n \sqrt{17} = \sqrt{m^{2} + 1}$ then $[n \sqrt{17}] = m$ for all $m, n \in \mathbb{Z}^{+}$.
Then we have $\{n \sqrt{17}\} = n \sqrt{17} - m$ which means
$$n \sqrt{17} - m > \frac{c}{n} \Leftrightarrow \frac{17 n^{2} - m^{2}}{n \sqrt{17} + m} > \frac{c}{n} \Leftrightarrow c < \frac{1}{\sqrt{17} + \frac{m}{n}}.$$
Note that $\frac{m}{n} = \sqrt{17 - \frac{1}{n^{2}}}$ then when $m, n \rightarrow +\infty$, we have $\frac{m}{n} \rightarrow \lim \sqrt{17 - \frac{1}{n^{2}}} = \sqrt{17}$. This implies that
$$c \leq \frac{1}{\sqrt{17} + \lim_{m, n \rightarrow +\infty} \frac{m}{n}} = \frac{1}{2 \sqrt{17}}$$
Hence, $c = \frac{1}{2 \sqrt{17}}$ is the maximum constant that satisfies the given condition.