Substitute the given values of P and Q into the expression (P)/(Q).

Numerator:

$$(4 + \sqrt{2})(2 - \sqrt{2}) = 4(2) - 4\sqrt{2} + 2\sqrt{2} - (\sqrt{2})^2$$ $$= 8 - 4\sqrt{2} + 2\sqrt{2} - 2$$ $$= 6 - 2\sqrt{2}$$

Denominator:

$$(2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - (\sqrt{2})^2$$ $$= 4 - 2$$ $$= 2$$

$$\frac{P}{Q} = \frac{6 - 2\sqrt{2}}{2}$$ $$= \frac{6}{2} - \frac{2\sqrt{2}}{2}$$ $$= 3 - \sqrt{2}$$

We have $3 - \sqrt{2} = a + b\sqrt{c}$.
By comparing the terms, we find:
$a = 3$
$b = -1$
$c = 2$

The values are $\boxed{a=3, b=-1, c=2}$.

2. Solving the trigonometric equation

$$4\sin^2 x + 4\cos x = 5$$ $$4(1 - \cos^2 x) + 4\cos x = 5$$

$$4 - 4\cos^2 x + 4\cos x = 5$$ $$-4\cos^2 x + 4\cos x - 1 = 0$$

Multiply by $-1$ to make the leading coefficient positive:

$$4\cos^2 x - 4\cos x + 1 = 0$$

This is a perfect square trinomial.

$$(2\cos x - 1)^2 = 0$$

Take the square root of both sides:

$$2\cos x - 1 = 0$$ $$2\cos x = 1$$ $$\cos x = \frac{1}{2}$$

The principal value for $\cos x = \frac{1}{2}$ is $x = 60^\circ$.
Since cosine is positive in the first and fourth quadrants, the other solution in the given range is:

$$x = 360^\circ - 60^\circ = 300^\circ$$

The solutions are $\boxed{x = 60^\circ, 300^\circ}$.

3. Making $a$ the subject of the formula

First, isolate the square root term by dividing by $k$.

$$\frac{R}{k} = \sqrt{y^2 - a^2}$$

$$\left(\frac{R}{k}\right)^2 = y^2 - a^2$$ $$\frac{R^2}{k^2} = y^2 - a^2$$

Add $a^2$ to both sides and subtract $\frac{R^2}{k^2}$ from both sides.

$$a^2 = y^2 - \frac{R^2}{k^2}$$

$$a = \pm\sqrt{y^2 - \frac{R^2}{k^2}}$$

The subject of the formula is $\boxed{a = \pm\sqrt{y^2 - \frac{R^2}{k^2}}}$.