Well this seems like $1-|t|$ for $|t|<1$ and $0$ for $|t|>1$ . Taking the Fourier transform $$X(ω) = \int_{-\infty}^\infty(1-|t|)e^{-jωt}dt\\=\int_{-\infty}^\infty e^{-jωt}dt -\int_{-\infty}^\infty|t|e^{-jωt}dt\\=2\piδ(ω)-\int_{-1}^1|t|e^{-jωt}dt\\=2πδ(ω)-\int_{-1}^0-te^{-jωt}dt-\int_{0}^1te^{-jωt}dt$$
The infinity integral goes from -1 to 1 since the function is zero elsewhere. Now I'm having trouble continuing with this. It doesn't seem to give me the result of the problem's solution which is $sinc^2(\frac{ω}{2\pi})$ , the sampling function.



The function is not $1-|t|$. That is only valid for $-1\le t\le 1$. Actually \begin{align} \int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt &=\int_{-1}^1(1-|t|)e^{-i\omega t}\,dt =2\int_0^1(1-t)\cos\omega t\,dt\\ &=\frac{2}{\omega}\left[(1-t)\sin\omega t\right]_0^1 +\frac{2}{\omega}\int_0^1\sin\omega t\,dt\\ &=-\frac2{\omega^2}\left[\cos\omega t\right]_0^1\\ &=\frac{2(1-\cos\omega)}{\omega^2}=\frac{4\sin^2(\omega/2)}{\omega^2}. \end{align}