(a)
Euler’s method: \( x_n = x_{n-1} + 0.1 \times x_{n-1} \cos t_{n-1} \times e^{-\sin t_{n-1}} \).
Initial condition: \( t_0 = 0 \), \( x_0 = \frac{1}{e} \approx 0.367879441 \).
Step 1: \( t_1 = 0.1 \), \( f(x_0, t_0) = 0.367879441 \times \cos 0 \times e^{-\sin 0} \approx 0.367879441 \), \( x_1 \approx 0.367879441 + 0.1 \times 0.367879441 \approx 0.404667385 \).
Step 2: \( t_2 = 0.2 \), \( f(x_1, t_1) \approx 0.404667385 \times \cos 0.1 \times e^{-\sin 0.1} \approx 0.404667385 \times 0.995004165 \times 0.904837418 \approx 0.364283902 \), \( x_2 \approx 0.404667385 + 0.1 \times 0.364283902 \approx 0.441095775 \).
Step 3: \( t_3 = 0.3 \), \( f(x_2, t_2) \approx 0.441095775 \times \cos 0.2 \times e^{-\sin 0.2} \approx 0.441095775 \times 0.980066578 \times 0.819801444 \approx 0.354456967 \), \( x_3 \approx 0.441095775 + 0.1 \times 0.354456967 \approx 0.476541472 \).
Round to 3 s.f.: \( x(0.3) \approx 0.477 \).

Result: \( x \approx 0.477 \) [4]

(b)
Solve: \( \frac{dx}{x} = \cos t \times e^{-\sin t} \, dt \).
Integrate: \( \ln |x| = \int \cos t \times e^{-\sin t} \, dt \), let \( u = -\sin t \), \( du = -\cos t \, dt \), so \( \int \cos t \times e^{-\sin t} \, dt = -e^u = -e^{-\sin t} + c \).
Thus: \( \ln |x| = -e^{-\sin t} + c \), \( x = A e^{-e^{-\sin t}} \).
Initial condition: \( t = 0 \), \( x = \frac{1}{e} \implies \ln \frac{1}{e} = -1 = -e^0 + c \implies c = 0 \).
Solution: \( x = e^{-e^{-\sin t}} \).
At \( t = 0.3 \): \( \sin 0.3 \approx 0.295520207 \), \( e^{-\sin 0.3} \approx 0.743937105 \), \( x \approx e^{-0.743937105} \approx 0.475139575 \).
Percentage error: \( \left| \frac{0.476541472 – 0.475139575}{0.475139575} \right| \times 100 \approx 0.296199\% \approx 0.296\% \).
Result: Percentage error \(\approx 0.296\%\) [4]