In right-angled triangle PQR, ∠Q = 90°. We are given that tan R = 1/2.

We can represent this in a right-angled triangle. Let QR = x and PQ = 2x. By Pythagoras' theorem, PR = √(PQ² + QR²) = √(4x² + x²) = x√5.

Now, let's find the trigonometric ratios:

• sin R = PQ/PR = 2x/(x√5) = 2/√5
• cos R = QR/PR = x/(x√5) = 1/√5
• sec R = PR/QR = √5
• cosec R = PR/PQ = √5/2
• sin P = QR/PR = x/(x√5) = 1/√5
• cos P = PQ/PR = 2x/(x√5) = 2/√5
• sec P = PR/PQ = √5

Substituting these values into the given expression:

\(\frac{\sec P (3 \cos R - \sin P)}{4 \csc R (4 \sin R - \cos P)} = \frac{\sqrt{5} (3(\frac{1}{\sqrt{5}}) - \frac{1}{\sqrt{5}})}{4(\frac{\sqrt{5}}{2})(4(\frac{2}{\sqrt{5}}) - \frac{2}{\sqrt{5}})} \)

\( = \frac{\sqrt{5} (\frac{2}{\sqrt{5}})}{4(\frac{\sqrt{5}}{2})(\frac{6}{\sqrt{5}})} = \frac{2}{2(6)} = \frac{2}{12} = \frac{1}{6} \)

There appears to be a discrepancy between the calculated value and the provided options. Let's re-examine the calculations. The expression simplifies to:

\( \frac{\sec P (3 \cos R - \sin P)}{4 \csc R (4 \sin R - \cos P)} = \frac{\sqrt{5} (3(1/\sqrt{5}) - 1/\sqrt{5})}{4(\sqrt{5}/2)(4(2/\sqrt{5}) - 2/\sqrt{5})} = \frac{\sqrt{5}(2/\sqrt{5})}{2\sqrt{5}(6/\sqrt{5})} = \frac{2}{12} = \frac{1}{6} \)

The provided options do not include 1/6. There might be an error in the question or the options.