$\cos\theta\cos2\theta\cos3\theta + \cos2\theta\cos3\theta\cos4\theta + ...$

$$\cos(n-1)t\cdot\cos nt\cdot\cos(n+1)t$$

$$=\dfrac{\cos nt(\cos2t+\cos2n t)}2$$

$$=\dfrac{\cos2t\cos nt}2+\dfrac{\cos nt+\cos3nt}4$$

$$=\dfrac{2\cos2t+1}4\cdot\cos nt+\dfrac{\cos3nt }4$$

Use $\sum \cos$ when angles are in arithmetic progression

Let's write $c_n:=\cos(n\theta),s_n:=\sin(n\theta)$ to simplify notation. We observe:$$ \begin{split}c_nc_{n+1}c_{n+2}&=c_{n+1}(c_{n+1}c_1+s_{n+1}s_1)(c_{n+1}c_1-s_{n+1}s_1)\\&=c_{n+1}(c_{n+1}^2c_1^2-s_{n+1}^2s_1^2)\\&=c_{n+1}(c_{n+1}^2(c_1^2+s_1^2)-s_1^2)\\&=c_{n+1}^3-c_{n+1}s_1.\end{split} $$ Hence, we see that $$\begin{split}\sum_{k=1}^nc_kc_{k+1}c_{k+2}=\sum_{k=2}^nc_k^3-s_1\sum_{k=2}^nc_k.\end{split}$$ You can find a reference for the latter sum here. For the first one, we may use the complex exponential representation $$\cos(\theta)=\frac12\left(e^{i\theta}+e^{-i\theta}\right)$$ in order to deduce that $$\cos(k\theta)^3=\frac18\left(e^{3ik\theta}+3e^{ik\theta}+3e^{-ik\theta}+e^{-3ik\theta}\right).$$

Summing this is now just a sum of four different geometric series, which should be easy to do (of course, you want to convert it back into sines and cosines at the end). This method can also be used for the second sum, actually.