Because this is for signal analysis class, I will assume that $\newcommand{\Res}{\operatorname{Res}}\newcommand{\sgn}{\operatorname{sgn}}\newcommand{\sinc}{\operatorname{sinc}}\sinc(t)=\frac{\sin(\pi t)}{\pi t}$.
Since $\sinc(4t)$ is an even function, we have
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin(4\pi t)}{4\pi t}e^{-2\pi i\,xt}\,\mathrm{d}t
&=\int_{-\infty}^\infty\frac{\sin(4\pi t)}{4\pi t}\cos(2\pi xt)\,\mathrm{d}t\\
&=\int_{-\infty}^\infty\frac{\sin(t(4\pi+2\pi x))+\sin(t(4\pi-2\pi x))}{8\pi t}\mathrm{d}t\tag{1}
\end{align}
$$
First note that $\int_{-\infty}^\infty\frac{\sin(k\pi t)}{\pi t}\mathrm{d}t$ is odd in $k$. For $k>0$, let's compute
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin(k\pi t)}{\pi t}\mathrm{d}t
&=\int_{-\infty}^\infty\frac{e^{i\pi kt}-e^{-i\pi kt}}{2\pi it}\mathrm{d}t\\
&=\int_{\gamma^+}\frac{e^{i\pi kt}}{2\pi it}\mathrm{d}t-\int_{\gamma^-}\frac{e^{-i\pi kt}}{2\pi it}\mathrm{d}t\\
&=2\pi i\Res\left(\frac{e^{i\pi kt}}{2\pi it},0\right)-0\\
&=2\pi i\cdot\frac{1}{2\pi i}\\
&=1\tag{2}
\end{align}
$$
where both $\gamma^+$ and $\gamma^-$ go from $-R-i$ to $R-i$ and then $\gamma^+$ circles back counter-clockwise to $-R-i$ and $\gamma^-$ circles back clockwise to $-R-i$ and $R\to\infty$.
$\hspace{4cm}$
Equation $(2)$ and its oddness in $k$ says that
$$
\int_{-\infty}^\infty\frac{\sin(k\pi t)}{\pi t}\mathrm{d}t=\sgn(k)\tag{3}
$$
Combining $(1)$ and $(3)$ yields
$$
\int_{-\infty}^\infty\frac{\sin(4\pi t)}{4\pi t}e^{-2\pi i\,xt}\,\mathrm{d}t=\frac{\sgn(4+2x)+\sgn(4-2x)}{8}\tag{4}
$$
which is $\frac14$ for $x\in\left(-2,2\right)$, $\frac18$ for $x\in\left\{-2,2\right\}$, and $0$ elsewhere.
