a. [6 marks]

METHOD 1:
Square both equations (M1):
\( (3\sin B + 4\cos C)^2 = 6^2 \implies 9\sin^2 B + 24\sin B \cos C + 16\cos^2 C = 36 \) (A1)
\( (4\sin C + 3\cos B)^2 = 1^2 \implies 16\sin^2 C + 24\cos B \sin C + 9\cos^2 B = 1 \) (A1)
Add equations and use \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( 9(\sin^2 B + \cos^2 B) + 16(\cos^2 C + \sin^2 C) + 24(\sin B \cos C + \cos B \sin C) = 36 + 1 \) (M1)
\( 9 + 16 + 24\sin (B + C) = 37 \implies 24\sin (B + C) = 12 \) (A1)
\( \sin (B + C) = \frac{12}{24} = \frac{1}{2} \) (A1) (AG)
Answer: \( \sin (B + C) = \frac{1}{2} \)

METHOD 2:
From \( 3\sin B + 4\cos C = 6 \), solve: \( \sin B = \frac{6 – 4\cos C}{3} \) (M1)
From \( 4\sin C + 3\cos B = 1 \), solve: \( \cos B = \frac{1 – 4\sin C}{3} \)
Use \( \sin (B + C) = \sin B \cos C + \cos B \sin C \):
\( \sin (B + C) = \left( \frac{6 – 4\cos C}{3} \right) \cos C + \left( \frac{1 – 4\sin C}{3} \right) \sin C = \frac{6\cos C + \sin C – 4}{3} \) (A1)
Similarly: \( \sin (B + C) = \sin B \left( \frac{6 – 3\sin B}{4} \right) + \cos B \left( \frac{1 – 3\cos B}{4} \right) = \frac{\cos B + 6\sin B – 3}{4} \) (M1, A1)
Add: \( 2\sin (B + C) = \frac{(18\sin B + 24\cos C) + (4\sin C + 3\cos B) – 25}{12} \) (A1)
\( = \frac{36 + 1 – 25}{24} = \frac{12}{24} \implies \sin (B + C) = \frac{1}{2} \) (A1) (AG)
Answer: \( \sin (B + C) = \frac{1}{2} \)

b. [5 marks]

In triangle \( ABC \), \( A + B + C = 180^\circ \), so \( \sin A = \sin (180^\circ – (B + C)) = \sin (B + C) \) (R1)
From part (a), \( \sin (B + C) = \frac{1}{2} \), so \( \sin A = \frac{1}{2} \) (A1)
Possible angles: \( A = 30^\circ \) or \( A = 150^\circ \) (A1)
If \( A = 150^\circ \), then \( B + C = 30^\circ \). Check: \( 3\sin B + 4\cos C = 6 \), but \( \sin B \leq 1 \), \( \cos C \leq 1 \), and for small \( B, C \), \( 3\sin B + 4\cos C < \frac{3}{2} + 4 = 5.5 < 6 \), a contradiction (R1)
Thus, only \( A = 30^\circ \) (when \( B + C = 150^\circ \)) is possible (R1) (AG)
Answer: \( \angle CAB = 30^\circ \) (only one value)