For the first one, observe that very "large" n, $\sqrt{n+c}\approx \sqrt n$ for any constant c. So the limit of $S_n$ is 0.
For the second one, divide numerator and denominator of $b_n$ by $(2/3)^n$ to get
$$b_n={1\over (3/4)^n+(27/20)^n}$$
I hope it is now clear that $b_n$ approaches 0.
Why can you not use the Squeeze Theorem?
For $S_n$, $6\sqrt{n-3}-6\sqrt{n+7} = 4\sqrt{n-3}-\sqrt{n+7}-5\sqrt{n+7}+2\sqrt{n-3} < S_n < 4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n-1}+2\sqrt{n+3} = 6\sqrt{n+3}-6\sqrt{n-1}$
$\displaystyle \begin{align*}\lim_{n \to \infty} \left(6\sqrt{n-3}-6\sqrt{n+7}\right) & = 6\lim_{n \to \infty} \left( \sqrt{n-3} - \sqrt{n+7} \right) \\ & = 6\lim_{n \to \infty} \dfrac{(n-3)-(n+7)}{\sqrt{n-3}+\sqrt{n+7}} \\ & = 6\lim_{n \to \infty} \dfrac{-10}{\sqrt{n-3} + \sqrt{n+7}} = 0\end{align*}$
Similarly for the upper bound. Apply the Squeeze Theorem.
Or do you mean the assignment says to not use the Squeeze Theorem?