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 $P = e^{\frac{9}{2}}$