Fourier Transform of shifted Impulse
Technically, the impulse $\delta(t)$ is called a distribution, and not a function, but for the purposes of your first course in Fourier transforms,  what you need to know is that $\delta(t)$ has the sifting property


  $$\displaystyle\int_{-\infty}^\infty x(t)\delta(t) \,\mathrm dt = x(0) ~\text{provided that }x(t) ~\text{is continuous at }t=0 \tag 1$$


which sifts the single value $x(0)$ out the function $x(t)$ and discards all other values. Also, any integral involving impulses can be manipulated via the change of variable technique into an integral of the form above to get the value of the integral. For example,

\begin{align}
\int_{-\infty}^\infty x(t)\delta(t-T) \,\mathrm dt
&amp;= \int_{-\infty}^\infty x(\tau+T)\delta(\tau) \,\mathrm d\tau
&amp;\scriptstyle{\text{set }\tau = t-T, ~ dt = d\tau}\\
&amp;= x(T)
\end{align}

and so we can think of $\delta(t-T)$ as a time-shifted impulse that plucks the value $x(T)$ (instead of $x(0)$) out of the function $x(t)$.
Similarly, for $a &gt;0$,
\begin{align}
\int_{-\infty}^\infty x(t)\delta(at) \,\mathrm dt
&amp;= \int_{-\infty}^\infty x\left(\frac{\tau}{a}\right)\delta(\tau) \,\frac 1a \mathrm d\tau
&amp;\scriptstyle{\text{set }\tau = at, ~ dt = \frac 1a d\tau}\\
&amp;= \frac 1a x(0)
\end{align}
Figure out why we can think of $\delta(at)$ as the magnitude-scaled impulse 
$\frac 1{|a|}\delta(t)$ and then apply both of the above ideas to
evaluating the Fourier transform integral
$$\int_{-\infty}^\infty e^{-j\omega t}\delta\left(\frac{t-T}{a}\right)
\, \mathrm dt$$