The diagram below shows the graphs of f(x) = 2sin 2x and g(x) = a tan bx. Give the coordinates of A and B. Write down the coordinates of C and D. What are the values of a and b?

Points A and B are the x-intercepts of the function $f(x) = 2\sin(2x)$. At x-intercepts, $f(x) = 0$.

$$2\sin(2x) = 0$$ $$\sin(2x) = 0$$

This occurs when $2x = n \cdot 180^\circ$ for any integer $n$.

$$x = n \cdot 90^\circ$$

From the graph, A is the first x-intercept to the left of the origin, so $n=-1$, which gives $x = -90^\circ$.
B is the first x-intercept to the right of the origin, so $n=1$, which gives $x = 90^\circ$.
The coordinates are:
A: $\boxed{(-90^\circ, 0)}$
B: $\boxed{(90^\circ, 0)}$

Point C is a local maximum of $f(x) = 2\sin(2x)$. The maximum value of $2\sin(2x)$ is 2.
This occurs when $\sin(2x) = 1$.

$$2x = 90^\circ + k \cdot 360^\circ$$ $$x = 45^\circ + k \cdot 180^\circ$$

From the graph, C is the first maximum in the positive x-axis, so $k=0$, which gives $x = 45^\circ$.
Point D is a local minimum of $f(x) = 2\sin(2x)$. The minimum value of $2\sin(2x)$ is -2.
This occurs when $\sin(2x) = -1$.

$$2x = -90^\circ + k \cdot 360^\circ$$ $$x = -45^\circ + k \cdot 180^\circ$$

From the graph, D is the first minimum in the negative x-axis, so $k=0$, which gives $x = -45^\circ$.
The coordinates are:
C: $\boxed{(45^\circ, 2)}$
D: $\boxed{(-45^\circ, -2)}$

The function is $g(x) = a\tan(bx)$.
From the graph, the vertical asymptotes of $g(x)$ are at $x = -90^\circ$ and $x = 90^\circ$.
For a tangent function, vertical asymptotes occur when the argument is $90^\circ + n \cdot 180^\circ$.
So, $bx = 90^\circ$.
Using $x = 90^\circ$:

$$b(90^\circ) = 90^\circ$$ $$b = 1$$

Now, use point C $(45^\circ, 2)$, which lies on $g(x)$. Substitute $x=45^\circ$, $y=2$, and $b=1$ into $g(x)$:

$$2 = a\tan(1 \cdot 45^\circ)$$ $$2 = a\tan(45^\circ)$$

Since $\tan(45^\circ) = 1$:

$$2 = a(1)$$ $$a = 2$$

The values are: $\boxed{a=2, b=1}$

We need to solve $2\sin(2x) > 2\tan(x)$ for $x \in [0^\circ; 90^\circ]$.
Divide by 2:

$$\sin(2x) > \tan(x)$$

Use the identities $\sin(2x) = 2\sin(x)\cos(x)$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$:

$$2\sin(x)\cos(x) > \frac{\sin(x)}{\cos(x)}$$

For $x \in (0^\circ, 90^\circ)$, both $\sin(x) > 0$ and $\cos(x) > 0$.
Multiply both sides by $\cos(x)$ (which is positive, so the inequality direction remains the same):

$$2\sin(x)\cos^2(x) > \sin(x)$$

Divide both sides by $\sin(x)$ (which is positive, so the inequality direction remains the same):

$$2\cos^2(x) > 1$$ $$\cos^2(x) > \frac{1}{2}$$

Since $\cos(x) > 0$ in the interval $(0^\circ, 90^\circ)$, take the positive square root:

$$\cos(x) > \sqrt{\frac{1}{2}}$$ $$\cos(x) > \frac{1}{\sqrt{2}}$$ $$\cos(x) > \frac{\sqrt{2}}{2}$$

We know that $\cos(45^\circ) = \frac{\sqrt{2}}{2}$. Since $\cos(x)$ is a decreasing function in the interval $(0^\circ, 90^\circ)$, for $\cos(x)$ to be greater than $\frac{\sqrt{2}}{2}$, $x$ must be less than $45^\circ$.
Also, at $x=0^\circ$, $f(0)=0$ and $g(0)=0$, so $f(0) \ngtr g(0)$.
At $x=45^\circ$, $f(45^\circ)=2$ and $g(45^\circ)=2$, so $f(45^\circ) \ngtr g(45^\circ)$.
Therefore, the inequality holds for $x$ values strictly between $0^\circ$ and $45^\circ$.
The values of $x$ are: $\boxed{0^\circ < x < 45^\circ}$