I already solved a problem for you related to this problem. You are right. The first limit is $0$. Here how to prove it. Making the change of variables $m=\ln(n)$ yields
$\lim_{n\to \infty}\frac{\log(n)^{\log(n)}}{1.01^{n}}= \lim_{m\to \infty}\frac{m^m}{e^{\ln(1.01)e^m}}=y\,.$
Taking the $\ln$ (the logarithmic function) to both sides of the last equation gives
$\implies \ln(y)=m\ln(m)-\ln(1.01)e^m \,,$
which follows from the properties of the logarithmic function. Taking the limit of the last equation gives
$ \implies \lim_{m\to \infty}\ln(y)= \ln(\lim_{m\to \infty}y) = \lim _{m\to \infty} (m\ln(m)-\ln(1.01)e^m)\rightarrow -\infty $
$ \implies \lim_{m\to \infty} y = e^{-\infty}=0 \,.$
Interchanging the order of the limit is justified by the continuity of the logarithmic function.