I don't fully follow what you are doing to determine the range. In any case, when you have:

$(1/y -3)^2\ge0$

The LHS is a square and thus always positive, this inequality is satisfied for all $y$...

In the formula: $$y = \frac{1}{3+\sqrt{x+1}}$$ the range of the monotonically increasing part $\sqrt{x+1}$ is (clearly) $[0,+\infty)$, which means the denominator is monotonically decreasing with a maximum in $x=-1$, namely $y = 1/3$. For $x \to \infty$, $y \to 0$ but since $y \ne 0$ for all $x$, the range is: $0 < y \le \tfrac{1}{3}$.

Alternatively: $$0 \le \sqrt{x+1} < +\infty$$ $$3 \le 3+\sqrt{x+1} < +\infty$$ $$ \frac{1}{3} \ge \frac{1}{3+\sqrt{x+1}} > \frac{1}{+\infty}$$ So: $$ 0 < \frac{1}{3+\sqrt{x+1}} \le \frac{1}{3}$$

For the domain of: $$g(x)=3+\sqrt{16-(x-3)^2}$$ You need: $$16-(x-3)^2 \ge 0 \iff (x-3)^2 \le 16 \iff |x-3| \le 4 \iff -1 \le x \le 7 $$ For the range (given the domain as above): $$0 \le 16-(x-3)^2 \le 16$$ $$0 \le \sqrt{16-(x-3)^2} \le 4$$ $$3 \le 3+ \sqrt{16-(x-3)^2} \le 7$$