For part a, you just have to compose the functions. Another way to write (f · g)(x) is f(g(x)), which means basically you take g(x) and let it be like a variable you plug into f(x). In your case, you look at f(x), and there isn't a variable there, so there's nothing you can compose because there's no x to plug g(x) into. (Unless you accidentally mistyped and there is supposed to be a variable in f(x). In that case, take g(x) and plug it in. For example, if f(x) = (1/2)x, (f · g)(x) would be (1/2)(x^2 + 5x).)
Anytime a question asks for domain and range, the first thing you want to ask yourself is if there's anything in either of your equations that limits the results. A pretty easy example of stuff that can "limit" a domain or range is something like 1/(x - 2), where x cannot equal 2, or else you'll be dividing by zero which will give you an error. Since there's nothing to compose with (f · g)(x) = 1/2, your domain will be (-∞, ∞) since it's a constant and when graphed, it'll extend forever. Your range is only one number, since 1/2 does not range above or below 1/2, but only sits exactly at 1/2. So your range is y = 1/2.
