求满足不等式组 $\dfrac{17-8x}{3} < \dfrac{2(3x-1)}{5} \le 4$ 的 $x$ 的取值范围。 由此，写出满足该不等式组的最小质数。

1. Solve the left-hand inequality

$\frac{17-8x}{3}<\frac{2(3x-1)}{5}$
Multiply both sides by $15$:
$5(17-8x)<6(3x-1)$
$85-40x<18x-6$
$91<58x$
$x>\frac{91}{58}$

2. Solve the right-hand inequality

$\frac{2(3x-1)}{5}\le 4$
Multiply both sides by $5$:
$2(3x-1)\le 20$
$6x-2\le 20$
$6x\le 22$
$x\le \frac{11}{3}$

3. Combine the two inequalities into the final range

From the two conditions,
$x>\frac{91}{58}\quad \text{and}\quad x\le \frac{11}{3}.$
Hence,
$\frac{91}{58}<x\le \frac{11}{3}.$

1. Identify the smallest prime number satisfying the range from part (a)

From part (a),
$1 \le x < \frac{79}{46}.$
Since $\frac{79}{46} \approx 1.72$, there is no prime number in this interval. The smallest prime number is $2$, but $2 \not< \frac{79}{46}$.

Therefore, no prime number satisfies the inequalities.