Three limits involving logarithm and exponentials

a) As $x\to0$, $e^{-x}\to1$, so $\ln(1+e^{-x})\to\ln2$. Hence
$$\lim_{x \to 0^-} \frac{\ln(1+e^{-x})}{x} = -\infty, \, \text{and} \, \lim_{x \to 0^+} \frac{\ln(1+e^{-x})}{x} = +\infty.$$
The limit does not exist.

b) As $x\to+\infty$, $e^{-x}\to0$, so $\ln(1+e^{-x})\to0$. The quotient is of the form $0/+\infty$, giving $0$.

c) Write $x^{x^x-1}=\exp\bigl((x^x-1)\ln x\bigr)$. As $x\to0^+$, $x(\ln x)^2\to0$ and
$$(x^x-1) \ln x = \frac{e^{x\ln x}-1}{x \ln x} \cdot x (\ln x)^2 \to 0.$$
By continuity of $\exp$, the limit is $e^0=1$.