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.