Fourier Transform with both Time Delay and Frequency Shift
If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:

$$\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)}\right\} &amp;= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0\left(t-t_0\right)} e^{-j2\pi f t}dt\\
\\
&amp;= \int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \left(\tau+t_0\right)}d\tau \\
\\
&amp;= e^{-j2\pi ft_0}\int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \tau}d\tau \\
\\
&amp;= e^{-j2\pi ft_0}\int_{-\infty}^\infty x\left(\tau\right) e^{-j2\pi (f-f_0) \tau}d\tau \\
\\
&amp;= e^{-j2\pi ft_0} X(f-f_0)\\
\\
\end{align*}$$

$$\begin{align*}\mathscr{F}\left\{x\left(t-t_0\right)e^{j2\pi f_0 t}\right\} &amp;= \int_{-\infty}^\infty x\left(t-t_0\right)e^{j2\pi f_0t} e^{-j2\pi f t}dt\\
\\
&amp;= \int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0(\tau+t_0)} e^{-j2\pi f \left(\tau+t_0\right)}d\tau \\
\\
&amp;= e^{-j2\pi (f-f_0)t_0}\int_{-\infty}^\infty x\left(\tau\right)e^{j2\pi f_0\tau} e^{-j2\pi f \tau}d\tau \\
\\
&amp;= e^{-j2\pi (f-f_0)t_0}\int_{-\infty}^\infty x\left(\tau\right) e^{-j2\pi (f-f_0) \tau}d\tau \\
\\
&amp;= e^{-j2\pi (f-f_0)t_0} X(f-f_0)\\
\\
\end{align*}$$