answer:We can factor the numerator to get y = frac{(x+1)(x+2)}{x+1}. If we exclude the case where x = -1, the function is equivalent to y = x+2. However, because x cannot equal -1, y cannot equal 1. Therefore, the range is all real numbers except for 1, which we may write as y in boxed{(-infty, 1)cup(1, infty)}.

answer:For the piecewise function to be continuous, the cases must "meet" at 2 and -2. For example, ax+3 and x-5 must be equal when x=2. This implies a(2)+3=2-5, which we solve to get 2a=-6 Rightarrow a=-3. Similarly, x-5 and 2x-b must be equal when x=-2. Substituting, we get -2-5=2(-2)-b, which implies b=3. So a+b=-3+3=boxed{0}.

answer:Notice that multiplying all three of the original equations together tells us that (a^2b^2c^2)/(abc) = 6, which implies abc=6. Rewriting the first and third equations as b = a^2 and c = sqrt{3a} and plugging these into abc=6 yields a cdot a^2cdot sqrt{3a} = 6. By squaring both sides of the equation, we obtain 3a^7 = 36 Rightarrow a = boxed{12^{1/7}}.

answer:Note that if the the two series have constant ratios of a and b, respectively, then 4left( frac{12}{1-a} right) = frac{12}{1-b}. Simplifying, 4(1-b)=1-a. Substituting in a= frac{4}{12}=frac{1}{3} and b= frac{4+n}{12}=frac{1}{3}+frac{n}{12}, we quickly find that n=boxed{6}.