We are given:
\( f(x) = e^{\cos 2x}. \)

First, we compute the derivative:
\( f'(x) = e^{\cos 2x} \cdot (-\sin 2x) \cdot 2. \)
\( f'(x) = -2 e^{\cos 2x} \sin 2x. \)

Setting \( f'(x) = 0 \):
\( -2 e^{\cos 2x} \sin 2x = 0. \)

Since \( e^{\cos 2x} \neq 0 \), we solve:
\( \sin 2x = 0. \)
\( 2x = k\pi, \quad k \in \mathbb{Z}. \)
\( x = \frac{k\pi}{2}. \)

Within the given domain:
\( -\frac{\pi}{4} \leq x \leq \frac{5\pi}{4}. \)

The valid solutions are:
\( x = 0, \quad x = \frac{\pi}{2}, \quad x = \pi. \)

Finding \( f(x) \) at these points:
\( f(0) = e^{\cos 0} = e^1 = e. \)
\( f\left(\frac{\pi}{2}\right) = e^{\cos \pi} = e^{-1} = \frac{1}{e}. \)
\( f(\pi) = e^{\cos 2\pi} = e^1 = e. \)

Thus, the points where the gradient is zero are:
\( (0, e), \quad \left(\frac{\pi}{2}, \frac{1}{e}\right), \quad (\pi, e). \)

Computing the second derivative:
\( f”(x) = -2 e^{\cos 2x} (2 \cos 2x \sin 2x) – 2 e^{\cos 2x} \cos 2x \cdot 2. \)
\( f”(x) = -4 e^{\cos 2x} \cos 2x \sin 2x – 4 e^{\cos 2x} \cos 2x. \)
\( f”(x) = -4 e^{\cos 2x} \cos 2x (1 + \sin 2x). \)

Evaluating at critical points:
At \( x = 0 \), \( \cos 2(0) = 1 \), \( \sin 2(0) = 0 \):
\( f”(0) = -4e (1)(1) = -4e < 0. \)
Local maximum at \( (0, e) \).

At \( x = \frac{\pi}{2} \), \( \cos \pi = -1 \), \( \sin \pi = 0 \):
\( f”\left(\frac{\pi}{2}\right) = -4e^{-1} (-1)(1) = 4/e > 0. \)
Local minimum at \( \left(\frac{\pi}{2}, \frac{1}{e}\right) \).

At \( x = \pi \), \( \cos 2\pi = 1 \), \( \sin 2\pi = 0 \):
\( f”(\pi) = -4e (1)(1) = -4e < 0. \)
Local maximum at \( (\pi, e) \).

\( f(x) \) has local maxima at \( (0, e) \) and \( (\pi, e) \).

\( f(x) \) has a local minimum at \( \left(\frac{\pi}{2}, \frac{1}{e}\right) \).

The curve is symmetric around \( x = \frac{\pi}{2} \).

(i) Maclaurin series for \( \cos 2x \):
\( \cos 2x = 1 – 2x^2 + \frac{4x^4}{3!} + O(x^6). \)

(ii) Maclaurin series for \( e^{\cos 2x – 1} \):
\( e^{\cos 2x – 1} = e^{(-2x^2 + \frac{4x^4}{3} + O(x^6))}. \)

Using \( e^y \approx 1 + y + \frac{y^2}{2} \) for small \( y \):
\( e^{-2x^2 + \frac{4x^4}{3}} \approx 1 – 2x^2 + \frac{4x^4}{3} + \frac{4x^4}{2} + O(x^6). \)
\( = 1 – 2x^2 + \frac{10x^4}{3} + O(x^6). \)

(iii) Maclaurin series for \( f(x) \):
\( f(x) = 1 – 2x^2 + \frac{10x^4}{3} + O(x^6). \)

\( \int_0^{1/10} e^{\cos 2x} \,dx \approx \int_0^{1/10} \left(1 – 2x^2 + \frac{10x^4}{3} \right) dx. \)

\( = \left[ x – \frac{2x^3}{3} + \frac{10x^5}{15} \right]_0^{1/10}. \)

\( = \left( \frac{1}{10} – \frac{2(1/10)^3}{3} + \frac{10(1/10)^5}{15} \right). \)

\( = \frac{1}{10} – \frac{2}{3000} + \frac{10}{150000}. \)

\( = \frac{1}{10} – \frac{1}{1500} + \frac{1}{15000}. \)

Approximating:
\( \approx \frac{149}{1500} e. \)

Thus,
\( \int_0^{1/10} e^{\cos 2x} dx \approx \frac{149e}{1500}. \)