In this answer we will prove that: $$\sum_{n=1}^\infty \lambda(n)\log\left(\cosh \left(\frac{1}{n}\right)\right)=\log\left(\frac{\cos\left(2\sqrt{\pi}\right)-\cosh\left(2\sqrt{\pi}\right)}{2\cos\left(\sqrt{2\pi}\right)-2\cosh\left(\sqrt{2\pi}\right)}\right),\ \ \ \ \ \ \ \ \ \ (1)$$
and
$$\sum_{n=1}^\infty \mu(n)\log \left(\cosh\left(\frac{1}{n}\right)\right)=\log \left(\frac{4+\pi^2}{1+\pi^2}\right).\ \ \ \ \ \ \ \ \ \ (2)$$
Outline: To prove this, we will first derive the series expansion for $\log\left(\cos\left(\frac{1}{n}\right)\right)$ which we obtain from the Weierstrass factorization for $\cos(x)$ (See this answer by Jack D'Aurizio's). We will insert this expansion into $\sum_{n=1}^\infty \lambda(n)\log(\cosh(1/n))$ and then change the order of summation (everything converges absolutely here). The inner sum over $n$ will have an explicit form due to the identity $$ \sum_{n=1}^\infty \frac{\lambda(n)}{n^s}=\frac{\zeta(2s)}{\zeta(s)}.\ \ \ \ \ \ \ \ \ \ (3)$$ Manipulating the resulting series, we will see that it can be decomposed as the sum of two series, each of which is of which can be written in terms of $\log(\sinh(\zeta_8 x))$ and $\log(\sin(\zeta_8 x))$ where $x=\sqrt{2\pi}$ or $x=\sqrt{\pi}$, and $\zeta_8$ is an $8^{th}$ root of unity. Finally, by rearranging this using identities for $\sin$ and $\sinh$, we obtain the final form that appears in equation $(1)$. Equation $(2)$ is much simpler, and follows from the same approach.
Lemma: We have the identities:
$$\log\left(\cosh(x)\right)-\sum_{k=1}^{\infty}\frac{4^{k}-1}{k\pi^{2k}}x^{2k}(-1)^{k}\zeta(2k)\ \ \ \ \ \ \ \ \ (4)$$
$$\log\left(\frac{\sin(x)}{x}\right)=-\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k\pi^{2k}}x^{2k}.\ \ \ \ \ \ \ \ \ (5)$$
$$\log\left(\frac{\sinh(x)}{x}\right)=-\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k\pi^{2k}}x^{2k}(-1)^{k}.\ \ \ \ \ \ \ \ \ (6)$$
Proof. Starting from the Weierstrass product for cosine and sine $$\cos x=\prod_{n=0}^{\infty}\left(1-\frac{4x^{2}}{(2n+1)^{2}\pi^{2}}\right),\ \ \ \sin x=z\prod_{n=0}^{\infty}\left(1-\frac{z^{2}}{n^{2}\pi^{2}}\right)$$ by taking logarithms and using the expansion $\log(1-z)=-\sum_{k=1}^\infty \frac{z^k}{k}$ we find that $$\log\left(\cos(x)\right)=\sum_{n=0}^{\infty}\log\left(1-\frac{4x^{2}}{(2n+1)^{2}\pi^{2}}\right)=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}-\frac{4^{k}x^{2k}}{k(2n+1)^{2k}\pi^{2k}}$$ $$=-\sum_{k=1}^{\infty}\frac{4^{k}x^{2k}}{k\pi^{2k}}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2k}}=-\sum_{k=1}^{\infty}\frac{4^{k}-1}{k\pi^{2k}}x^{2k}\zeta(2k)$$ since $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{s}}=\sum_{n=1}^\infty
\frac{1}{n^s}-\sum_{n=1}^\infty
\frac{1}{(2n)^s}=(1-2^{-s})\zeta(s).$$ Similarly, $$\log\left(\frac{\sin(x)}{x}\right)=\sum_{n=0}^{\infty}\log\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)=-\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k\pi^{2k}}x^{2k}.$$ The lemma then follows from the identities $$\cos(ix)=\cosh\left(x\right),\ \ \ \ \ \frac{\sin(ix)}{ix}=\frac{\sinh(x)}{x}.$$
Proof of equation $(1)$:
Using $(4)$, the series expansion for $\log\left(\cosh\left(\frac{1}{n}\right)\right),$ we have that $$\sum_{n=1}^{\infty}\lambda(n)\log\left(\cosh\left(\frac{1}{n}\right)\right)=-\sum_{k=1}^{\infty}\frac{4^{k}-1}{k\pi^{2k}}(-1)^{k}\zeta(2k)\sum_{n=1}^{\infty}\frac{\lambda(n)}{n^{2k}}$$ and so by $(3)$ we have that $$\sum_{n=1}^{\infty}\lambda(n)\log\left(\cosh\left(\frac{1}{n}\right)\right)=-\sum_{k=1}^{\infty}\frac{4^{k}-1}{k\pi^{2k}}(-1)^{k}\zeta(4k).$$
Splitting up the numerator and rearranging, this equals $$-\sum_{k=1}^{\infty}\frac{\zeta(4k)}{\pi^{4k}k}(-1)^{k}\left(2\pi\right)^{2k}+\sum_{k=1}^{\infty}\frac{\zeta(4k)}{\pi^{4k}k}(-1)^{k}\pi^{2k}.\ \ \ \ \ \ \ \ \ \ (7)$$ Define $$F(x)=-\sum_{k=1}^{\infty}\frac{\zeta(4k)}{k\pi^{4k}}x^{4k},$$ and notice that by adding equations $(5)$ and $(6)$ we have that $$F(x)=\log\left(\frac{\sin(x)}{x}\right)+\log\left(\frac{\sinh(x)}{x}\right).$$ Let $\zeta_{8}=\frac{1+i}{\sqrt{2}}$, which is an $8^{th}$ root of unity and satisfies $\zeta_{8}^{4}=-1$. Then equation $(7)$ along with the definition of $F(x)$ tells us that $$\sum_{n=1}^{\infty}\lambda(n)\log\left(\cosh\left(\frac{1}{n}\right)\right)=F\left(\zeta_{8}\sqrt{2\pi}\right)-F\left(\zeta_{8}\sqrt{\pi}\right).$$ Hence, our series is equal to $$\log\left(\frac{\sin\left((1+i)\sqrt{\pi}\right)}{(1+i)\sqrt{\pi}}\right)+\log\left(\frac{\sinh\left((1+i)\sqrt{\pi}\right)}{(1+i)\sqrt{\pi}}\right)-\log\left(\frac{\sin\left(\frac{1+i}{\sqrt{2}}\sqrt{\pi}\right)}{\frac{1+i}{\sqrt{2}}\sqrt{\pi}}\right)-\log\left(\frac{\sinh\left(\frac{1+i}{\sqrt{2}}\sqrt{\pi}\right)}{\frac{1+i}{\sqrt{2}}\sqrt{\pi}}\right),$$ and this becomes $$\log\left(\frac{1}{2}\cdot\frac{\sin\left((1+i)\sqrt{\pi}\right)}{\sin\left(\frac{1+i}{\sqrt{2}}\sqrt{\pi}\right)}\cdot\frac{\sinh\left((1+i)\sqrt{\pi}\right)}{\sinh\left(\frac{1+i}{\sqrt{2}}\sqrt{\pi}\right)}\right).$$ By writing $\sinh$ and $\sin$ as sums of exponentials, we have that $$\sin\left((1+i)x\right)\sinh\left((1+i)x\right)=\frac{e^{i(1+i)x}-e^{-i(1+i)x}}{2i}\cdot\frac{e^{(1+i)x}-e^{-(1+i)x}}{2}$$
$$=\frac{e^{-(1-i)x}-e^{(1-i)x}}{2i}\cdot\frac{e^{(1+i)x}-e^{-(1+i)x}}{2}=\frac{1}{4i}\left(e^{2ix}+e^{-2ix}-e^{2x}-e^{-2x}\right),$$ and so $$\sin\left((1+i)x\right)\sinh\left((1+i)x\right)=\frac{1}{2i}\left(\cos(2x)-\cosh(2x)\right).$$
From this we obtain the desired result:
$$\sum_{n=1}^{\infty}\lambda(n)\log\left(\cosh\left(\frac{1}{n}\right)\right)=\log\left(\frac{\cos\left(2\sqrt{\pi}\right)-\cosh\left(2\sqrt{\pi}\right)}{2\cos(\sqrt{2\pi})-2\cosh(\sqrt{2\pi)}}\right).$$
Proof of equation $(2)$:
Since $$\sum_{n=1}^\infty \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)},$$ by using the expansion $(4)$ we have that $$\sum_{n=1}^\infty \mu(n)\log\left(\cosh\left(\frac{1}{n}\right)\right)=-\sum_{k=1}^{\infty}\frac{4^{k}-1}{k\pi^{2k}}(-1)^{k}.$$ This equals $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\frac{2^{2k}}{\pi^{2k}}-\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\frac{1}{\pi^{2k}},$$ and since $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} x^k$, we see that
$$\sum_{n=1}^\infty \mu(n)\log\left(\cosh\left(\frac{1}{n}\right)\right)=\log \left(1+\frac{4}{\pi^2}\right)-\log \left(1+\frac{1}{\pi^2}\right)=\log \left(\frac{4+\pi^2}{1+\pi^2}\right).$$
