Let’s go through the problems step by step.

1. Express each of the following as a single trigonometric ratio:

(a) $2 \sin 5x \cos 5x$
Using the double-angle formula:
$2 \sin A \cos A = \sin(2A)$
$\therefore 2 \sin 5x \cos 5x = \sin(10x)$

(b) $\cos^2 \theta - \sin^2 \theta$
Using the cosine double-angle identity:
$\cos^2 A - \sin^2 A = \cos(2A)$
$\therefore \cos^2 \theta - \sin^2 \theta = \cos(2\theta)$

(c) $1 - 2 \sin^2 3x$
Using the cosine double-angle identity:
$1 - 2 \sin^2 A = \cos(2A)$
$\therefore 1 - 2 \sin^2 3x = \cos(6x)$

(d) $\frac{2 \tan 4x}{1 - \tan^2 4x}$
Using the tangent double-angle formula:
$\frac{2 \tan A}{1 - \tan^2 A} = \tan(2A)$
$\therefore \frac{2 \tan 4x}{1 - \tan^2 4x} = \tan(8x)$

(e) $4 \sin \theta \cos \theta$
Using the double-angle formula:
$2 \sin A \cos A = \sin(2A)$
$4 \sin \theta \cos \theta = 2 \sin(2\theta)$

(f) $2 \cos^2 \frac{\theta}{2} - 1$
Using the cosine double-angle identity:
$2 \cos^2 A - 1 = \cos(2A)$
$\therefore 2 \cos^2 \frac{\theta}{2} - 1 = \cos(\theta)$

2. Express each of the following as a single trigonometric ratio and evaluate:

(a) $2 \sin 45^\circ \cos 45^\circ$
Using $2 \sin A \cos A = \sin(2A)$:
$2 \sin 45^\circ \cos 45^\circ = \sin(90^\circ) = 1$

(b) $\cos^2 30^\circ - \sin^2 30^\circ$
Using $\cos^2 A - \sin^2 A = \cos(2A)$:
$\cos^2 30^\circ - \sin^2 30^\circ = \cos(60^\circ) = \frac{1}{2}$

(c) $2 \sin \frac{\pi}{12} \cos \frac{\pi}{12}$
Using $2 \sin A \cos A = \sin(2A)$:
$2 \sin \frac{\pi}{12} \cos \frac{\pi}{12} = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

(d) $\cos^2 \frac{\pi}{12} - \sin^2 \frac{\pi}{12}$
Using $\cos^2 A - \sin^2 A = \cos(2A)$:
$\cos^2 \frac{\pi}{12} - \sin^2 \frac{\pi}{12} = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

(e) $1 - 2 \sin^2 \frac{3\pi}{8}$
Using $1 - 2 \sin^2 A = \cos(2A)$:
$1 - 2 \sin^2 \frac{3\pi}{8} = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

(f) $2 \tan 60^\circ \cos^2 60^\circ$
Evaluate step by step:
[ \tan 60^\circ = \sqrt{3}, \cos^2