= Solution
Integrating the interval indicator gives
$$
\boxed{\hat f_1(\lambda)=\frac{2\sin(\lambda/2)}{\lambda},\quad\hat f_1(0)=1.}
$$
The <convolution> $f_2(x)$ is the length of overlap of two intervals of length one whose centres differ by $x$. Thus $f_2(x)=1-|x|$ for $|x|\leq1$, and zero otherwise. The <convolution theorem> gives
$$
\boxed{\hat f_2(\lambda)=\left(\frac{2\sin(\lambda/2)}{\lambda}\right)^2.}
$$
For this <Fourier transform> normalization, the <Parseval identity> says $\int|f|^2dx=(2\pi)^{-1}\int|\hat f|^2d\lambda$. Since
$$
\int|f_2(x)|^2dx=2\int_0^1(1-x)^2dx=\frac23,
$$
we have $\int|\hat f_2|^2d\lambda=4\pi/3$. Substituting $\lambda=2y$ gives
$$
2\int_{-\infty}^{\infty}\left(\frac{\sin y}{y}\right)^4dy=\frac{4\pi}{3},
\qquad\boxed{\int_{-\infty}^{\infty}\left(\frac{\sin y}{y}\right)^4dy=\frac{2\pi}{3}.}
$$
The quotient at zero is understood by its continuous limit.
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