To solve the problem, let's start by analyzing the given equation: \(4\sin\theta + \csc\theta = 4\), where \(0^\circ < \theta < 90^\circ\).

1. Recall that \(\csc\theta = \frac{1}{\sin\theta}\). So, substitute this into the equation: \(4\sin\theta + \frac{1}{\sin\theta} = 4\).
2. Let \(\sin\theta = x\) for simplicity. Then the equation becomes: \(4x + \frac{1}{x} = 4\).
3. Multiply throughout by \(x\) to eliminate the fraction: \(4x^2 + 1 = 4x\).
4. Rearrange to form a quadratic equation: \(4x^2 - 4x + 1 = 0\).
5. Use the quadratic formula to solve for \(x\), \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -4\), and \(c = 1\): \(x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}\)
6. Simplify under the square root: \(x = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4 \pm 0}{8} = \frac{4}{8} = \frac{1}{2}\)
7. Thus, \(\sin\theta = \frac{1}{2}\). In the first quadrant, this implies \(\theta = 30^\circ\).
8. Now, let's find \(\sin 3\theta\) and \(\cos 3\theta\). Use the identities:
   • \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\)
   • \(\cos 3\theta = 4\cos^3\theta - 3\cos\theta\)
9. Since \(\sin\theta = \frac{1}{2}\), calculate \(\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}\).
10. Compute \(\sin 3\theta\): \(\sin 3\theta = 3 \times \frac{1}{2} - 4 \times \left(\frac{1}{2}\right)^3 = \frac{3}{2} - \frac{4}{8} = \frac{3}{2} - \frac{1}{2} = 1\).
11. Compute \(\cos 3\theta\): \(\cos 3\theta = 4 \times \left(\frac{\sqrt{3}}{2}\right)^3 - 3 \times \frac{\sqrt{3}}{2}\)
12. Simplify \(\cos 3\theta\): \(= 4 \times \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = 0\)
13. Add these results: \(\sin 3\theta + \cos 3\theta = 1 + 0 = 1\).

Therefore, the value of \(\sin 3\theta + \cos 3\theta\) is 1.