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I know that the Fourier transform of a function with time delay can be written as: $$mathscrFbigx(t-t_0)big=X(f)e^-j2pi f t_0$$

The Fourier transform of a function with frequency shift can also be written as: $$mathscrFBigx(t)e^j2pi f_0 tBig=X(f-f_0)$$

So what if we have both shift and delay at the time domain, what will be the result in the frequency domain? E.g.:

$$mathscrFBigx(t-t_0)e^j2pi f_0 (t-t_0)Big$$

Will the result be: $$X(f-f_0)e^-j 2 pi f (t-t_0)$$

Also what will be the result of:

$$mathscrFBigx(t-t_0)e^j2pi f_0 t)Big$$

Is there an order to apply these properties?

As an alternative to going back to the definitions, as explained in Andy Walls' answer, you can also just apply the rules as you stated them:

$$mathcalFleftx(t-t_0)e^j2pi f_0(t-t_0)right=mathcalFleftx(t)e^j2pi f_0trighte^-j2pi ft_0=X(f-f_0)e^-j2pi ft_0$$

where $X(f)$ is the Fourier transform of $x(t)$.

And, for your second example, with $tildeX(f)=mathcalFx(t-t_0)=X(f)e^-j2pi ft_0$ you get

$$mathcalFleftx(t-t_0)e^j2pi f_0tright=tildeX(f-f_0)=X(f-f_0)e^-j2pi (f-f_0)t_0$$

Of course, recognizing that the function in the second example just equals the first function scaled by $e^j2pi f_0t_0$, we could have written down its Fourier transform directly by scaling the Fourier transform of the first function by the same factor.

If you are ever unsure, just go back to the definition and work out the Fourier Transform property for the specific situation:

$$beginalign*mathscrFleftxleft(t-t_0right)e^j2pi f_0left(t-t_0right)right &= int_-infty^infty xleft(t-t_0right)e^j2pi f_0left(t-t_0right) e^-j2pi f tdt\
\
&= int_-infty^infty xleft(tauright)e^j2pi f_0tau e^-j2pi f left(tau+t_0right)dtau \
\
&= e^-j2pi ft_0int_-infty^infty xleft(tauright)e^j2pi f_0tau e^-j2pi f taudtau \
\
&= e^-j2pi ft_0int_-infty^infty xleft(tauright) e^-j2pi (f-f_0) taudtau \
\
&= e^-j2pi ft_0 X(f-f_0)\
\
endalign*$$

$$beginalign*mathscrFleftxleft(t-t_0right)e^j2pi f_0 tright &= int_-infty^infty xleft(t-t_0right)e^j2pi f_0t e^-j2pi f tdt\
\
&= int_-infty^infty xleft(tauright)e^j2pi f_0(tau+t_0) e^-j2pi f left(tau+t_0right)dtau \
\
&= e^-j2pi (f-f_0)t_0int_-infty^infty xleft(tauright)e^j2pi f_0tau e^-j2pi f taudtau \
\
&= e^-j2pi (f-f_0)t_0int_-infty^infty xleft(tauright) e^-j2pi (f-f_0) taudtau \
\
&= e^-j2pi (f-f_0)t_0 X(f-f_0)\
\
endalign*$$