Given P = 4 + sqrt(2) and Q = 2 + sqrt(2).

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:

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

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

$$3 - \sqrt{2} = a + b\sqrt{c}$$

By comparison, we have:
$a = 3$
$b = -1$
$c = 2$

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

---

:
Solve the equation $4\sin^2 x + 4\cos x = 5$ for $0^\circ \le x \le 360^\circ$.

Substitute this into the equation.

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

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

Multiply by -1 to make the leading coefficient positive:

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

The equation becomes a quadratic equation in $y$:

$$4y^2 - 4y + 1 = 0$$

This is a perfect square trinomial: $(2y - 1)^2 = 0$.
Alternatively, use the quadratic formula: $y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)}$

$$y = \frac{4 \pm \sqrt{16 - 16}}{8}$$ $$y = \frac{4 \pm 0}{8}$$ $$y = \frac{4}{8}$$ $$y = \frac{1}{2}$$

The cosine function is positive in the first and fourth quadrants.
The reference angle is $\cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$.

In the first quadrant:

$$x = 60^\circ$$

In the fourth quadrant:

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

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

---

:
Make $a$ the subject of the formula: $R = h\sqrt{y^2 - a^2}$.

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

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

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

Remember to include both positive and negative roots.

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

This can also be written with a common denominator inside the square root:

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

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