There is a simple way to solve $1^\infty$ type limits. Suppose you need to find $\lim f(x)^{g(x)}, f(x) \to 1, g(x)\to \infty$
Let $P = f(x)^{g(x)}$ Define $p = \log{P} = g(x)\log{f(x)}$ As $f$ is close to one, its logarithm can be expanded using the standard Taylor series for logs. $\log(1+x) = x -\frac{x^2}{2}+\ldots$ So, $\log(f(x))\approx f(x)-1 - \frac{(f(x)-1)^2}{2}+\ldots$ Usually the first term is enough. So, $p\approx g(x)(f(x)-1) + \mbox{higher order terms}$ So, $P = \lim_{x\to x_0} e^{g(x)(f(x)-1)}$
Applying it to your case, 1. $p = g(x)(f(x)-1) = (x^2 + \frac{8}{x})(\frac{1}{9x^2+\ldots})$ $\lim_{x \to \infty} p = \frac{1}{9}$ So, your final answer is $P = e^p = e^{\frac{1}{9}}$
For 2, you can tell the answer by inspection. $P = e^{\frac{8}{9}}$
For 3, you need $p = \lim_{x \to 0}\frac{\sqrt{1+9x}-1}{x}$ Binomial expansion or L'Hospital Rule gives $p = \frac{9}{2}$ So, your answer is $P = e^{\frac{9}{2}}$