To find (a) and (b), use the difinitions of cscx=1/sinx and tanx=sinx/cosx.
For (c) and (d) first clarify which f and g do you mean and then use definition of composition of functions (f∘g)(x)=f(g(x)) and subsituate g(x) instead of x in f. Similar for g∘f.
Domain and range for which ones?
$\displaystyle f(x) = \dfrac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}}$
$\displaystyle g(x) = \dfrac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}}$
For (a), just multiply the two functions together:
$\displaystyle (f \cdot g)(x) = \left( \dfrac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) \left( \dfrac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}} \right)$
For (b), divide the two functions together (which is the same as multiplying by the reciprocal):
$\displaystyle (f/g)(x) = \left( \dfrac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) \left( \dfrac{1 - \sqrt[3]{x + 1}}{2x - \sqrt{x}} \right)$
For (c), in f(x), plug in g(x) for x:
$\displaystyle (f \circ g)(x) = \dfrac{6 \left( \sqrt{\frac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}}} \right) + \left( \frac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}} \right)^2}{\sqrt[5]{\left( \frac{2x - \sqrt{x}}{1 - \sqrt[3]{x + 1}} \right) - 2}}$
For (d), in g(x), plug in f(x) for x:
$\displaystyle (g \circ f)(x) = \dfrac{2 \left( \frac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) - \sqrt{\frac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}}}}{1 - \sqrt[3]{\left( \frac{6\sqrt{x} + x^2}{\sqrt[5]{x - 2}} \right) + 1}}$
I hope that you don't have to simplify these.