For the first one, you can apply the quotient rule. You can use numeric exponents (instead of square roots). Before you differentiate:

$y = \dfrac{\sqrt{x}-2}{\sqrt{x}+2} = \dfrac{x^{1/2}-2}{x^{1/2}+2}$

Next, notice that if you add 2 and subtract 2 in the numerator, you get the denominator at least once:

$y = \dfrac{(x^{1/2}+2)-4}{x^{1/2}+2} = \dfrac{x^{1/2}+2}{x^{1/2}+2} - \dfrac{4}{x^{1/2}+2} = 1-4(x^{1/2}+2)^{-1}$

Do you know how to find the derivative of either of these two? I have not used any calculus yet, just algebra, so everything I just did should be straightforward.

For $F(x) = (x^5-1)^4(2-3x)^3$, you need to use the product rule and the chain rule multiple times. Let $g=(x^5-1)^4$ and $h=(2-3x)^3$. Then $F(x) = g\cdot h$. Taking the derivative using the product rule, you find $F'(x) = \dfrac{dg}{dx}h + g\dfrac{dh}{dx}$. To find $\dfrac{dg}{dx}$ and $\dfrac{dh}{dx}$, you can let $j = x^5-1$ and $k=2-3x$. This gives you $g=j^4$ and $h=k^3$. So, $\dfrac{dg}{dx} = 4j^3\dfrac{dj}{dx}$ and $\dfrac{dh}{dx} = 3k^2\dfrac{dk}{dx}$. You should be able to easily calculate both $\dfrac{dj}{dx}$ and $\dfrac{dk}{dx}$ and then just plug in what you find.