1. I would like to know if this is correct:

$\displaystyle 5sinx-2secx=0$ where x is an element of 0 and 360 (2pi in radians)

$\displaystyle 5sinx-\frac {2}{cosx}=0$

$\displaystyle 5sinxcosx-2=0$

$\displaystyle sin5x-2=0$

No. 5 sin x cos x = 2.5 sin 2x.
$\displaystyle \therefore sinx=\frac {2}{3}, \frac {2}{3} + \pi, \frac {2}{3} + 2\pi$

2. I would like to know where to continue in this question:

$\displaystyle sin3xcosx-cos3xsinx=\frac {\sqrt{3}}{2}$

$\displaystyle 3sinx-4cos^3xcosx-4cos^3x-3cosx=$

Don't do it like this. Use sin 3x cos x - cos 3x sin x = sin (3x-x) = sin 2x
$\displaystyle 3sinxcosx-4cos^4x-4sincos^3x-3cos^2xsinx=$

...I don't really know where to go from here and how to find angles for x.

3. I am being introduced to product and sum formulas and I don't know how to tackle the questions.

eg1. $\displaystyle sinx=\frac{3}{2}, 0<x<\frac{\pi}{2}$

This doesn't make sense. Do you mean tan x = 3/2?
Find the exact value of sin2x, cos2x.

The extra formulas given for sinx and cosx are:

$\displaystyle sinx=\frac{2\color{red}\tan\color{black}\frac{x}{2 }}{1+tan^2 \frac{x}{2}}$

and

$\displaystyle cosx=\frac {1-tan^2 \frac{x}{2}}{1+tan^2\frac{x}{2}}$

I know all the normal ones like sin2x=2sinxcosx etc., just not how to apply them.

Don't use them here. Evaluate sin 2x and cos 2x, using the formula above, but substituting 2x in place of x.
If you could help me out here that would be great