The Fourier transform of sinusoids' products with possible other components
HINT:
It is easy to see that things can be simplified using the trigonometric product-to-sum identity in Equation $(1)$ below:
$$
\cos(\alpha)\cos(\beta) = \frac 12\big[\cos(\alpha+\beta) + \cos(\alpha-\beta)\big]\tag{1}
$$

In the first example
$$
\frac{1}{t^2}\cdot\cos(2πt)\cdot\cos(2πt) = \frac 1{2t^2}\big(\cos(4\pi t) + 1\big)
$$
From there you would want to visit the cosine modulation frequency-shift and differentiation properties of the Fourier transform, here in $(2)$ and $(3)$ respectively.
\begin{align}
\mathcal F\big\{x(t)\cos(2\pi f_0 t)\big\} &amp;= \frac 12\big[X(f - f_0) + X(f + f_0)\big]\tag{2}\\
\mathcal F\left\{\frac{d^n x(t)}{dt^n}\right\}&amp;=  \left(j2\pi f\right)^nX(f)\tag{3}
\end{align}

In the second example, using Equation $(1)$ you then have
\begin{align}
8\cos(20\pi t)\cos(40\pi t)\cos(80\pi t) &amp; = 8\bigg(\frac 12\big(\cos(60\pi t) + \cos(20\pi t)\big)\cos(80\pi t)\bigg)\\
&amp; = 4\big(\cos(60\pi t) + \cos(20\pi t)\big)\cos(80\pi t)\\
&amp; = 2\big(\cos(140\pi t) + \cos(20\pi t)\big)\\
&amp;\quad + 2\big(\cos(100\pi t) + \cos(60\pi t)\big)\\
\end{align}
With this you simply have sums and you don't have to think of convolutions, you have individual sinusoids at frequencies $10\ \rm Hz$, $30\ \rm Hz$, $50\ \rm Hz$, and $70\ \rm Hz$.