= Solution
Regard the same triangular functions as cutoffs in the frequency variable. They satisfy
$$
0\leq\theta_n(\xi)\uparrow1.
$$
Since $\widehat f\geq0$, monotone convergence gives
$$
\|\widehat f\|_1
=\lim_{n\to\infty}\int_{\mathbb R}\theta_n(\xi)\widehat f(\xi)\,d\xi.
$$
Fubini's theorem and evenness of $\theta_n$ give
$$
\int\theta_n(\xi)\widehat f(\xi)\,d\xi
=\int f(x)\widehat\theta_n(x)\,dx.
$$
Part (b) shows that $\widehat\theta_n\geq0$. Moreover, Fourier inversion at zero yields
$$
\int_{\mathbb R}\widehat\theta_n(x)\,dx
=2\pi\theta_n(0)=2\pi.
$$
Consequently
$$
0\leq\int\theta_n\widehat f
=\left|\int f\widehat\theta_n\right|
\leq\|f\|_\infty\int\widehat\theta_n
=2\pi\|f\|_\infty.
$$
Passing to the limit proves
$$
\boxed{\|\widehat f\|_{L^1}\leq2\pi\|f\|_{L^\infty}.}
$$
\b[Thus one may take the universal constant $\alpha=2\pi$.]
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