Newton's Law of cooling problem when the thermometer brought indoors
I will use $T(t)=T_m+(T_0-T_m)e^{kt}$ for both case.
Outdoor:
You done correctly,$$T_{\text{outdoor}}(t)=60e^{kt}+20\quad\text{ Where }k=\frac{1}{3}\ln\left(\frac{11}{33}\right)$$
Indoor:
Here you are also correct. I will just change your notation taking $\color{green}{T_0}=\underbrace{60e^{kt}+20}_\color{red}{{T_{\text{outdoor}}(t)}}$. Then $T_{\text{indoor}}(t)=80+(\color{green}{T_0}-80)e^{kt}$. Suppose after $t_1$ min. the thermometer brought indoor then $T_{\text{indoor}}(10-t_1)=71^{\circ}F$ where $k=\frac{1}{3}\ln\left(\frac{11}{33}\right)$
\begin{align}
T_{\text{indoor}}(10-t_1)&=71\\
80+(60e^{kt_1}+20-80)e^{k(10-t_1)}&=71\\
\vdots\\
t\approx 4.95
\end{align}
Hence the thermometer brought indoors at $2:05$PM