1.) By factoring, you can get this limit to the form:
$$L=a\cdot\lim_{n\to\infty}b^n$$
where $0<a,b\in\mathbb{R}$.
At this point, you may divide through by $a$ to obtain:
$$\frac{L}{a}=\lim_{n\to\infty}b^n$$
Next, take the natural log of both sides, and apply the property of limits:
$$\log_a\left(\lim_{x\to c}f(x) \right)=\lim_{x\to c}\left(\log_a\left(f(x) \right) \right)$$
to obtain:
$$\ln\left(\frac{L}{a} \right)=\lim_{n\to\infty}\ln\left(b^n \right)$$
Using the log property $$\log_a\left(b^c \right)=c\cdot\log_a(b)$$ we may write:
$$\ln\left(\frac{L}{a} \right)=\lim_{n\to\infty}n\cdot\ln\left(b \right)$$
$$\ln\left(\frac{L}{a} \right)=\ln\left(b \right)\lim_{n\to\infty}n$$
If $0<b<1$, we have:
$$\ln\left(\frac{L}{a} \right)=-\infty$$
Converting from logarithmic to exponential form, we have:
$$\frac{L}{a}=e^{-\infty}=0\implies L=0$$
If $b=1$ then we have:
$$\ln\left(\frac{L}{a} \right)=0$$
Converting from logarithmic to exponential form:
$$\frac{L}{a}=1\implies L=a$$
If $1<b$ then we have:
$$\ln\left(\frac{L}{a} \right)=\infty$$
Converting from logarithmic to exponential form, we find:
$$\frac{L}{a}=e^{\infty}=\infty\implies L=\infty$$
What do you find?