Your sum can be re-written in terms of an Euler product:
$\sum_{a=1}^\infty\sum_{b=1}^\infty\sum_{c=1}^\infty\sum_{d=1}^\infty\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^2b^2c^2d^2}\gcd(a,b,c,d)^4=\frac{1296}{\pi^8}\prod_{p}(1+\frac{p^4-1}{(p^2-1)^4})\approx .16544\cdots$
A proof can be given as follows:
First note that,
$d\mid x_1\wedge d\mid x_2\wedge d\mid x_3\wedge d\mid x_4\iff d\mid \gcd(x_1,x_2,x_3,x_4)$
So we get:$\sum_{d=1}^\infty f(d)1_{d\mid x_1}1_{d\mid x_2}1_{d\mid x_3}1_{d\mid x_4}=\sum_{d=1}^\infty f(d)1_{d\mid x_1\wedge d\mid x_2\wedge d\mid x_3\wedge d\mid x_4}=\sum_{d=1}^\infty f(d)1_{d\mid \gcd(x_1,x_2,x_3,x_4)}$ $=\sum_{d\mid \gcd(x_1,x_2,x_3,x_4)}f(d)$
Where $1_{A}=[A]$ in Iverson bracket notation
Then define:
$\phi_s(n)=n^s\prod_{p\mid n}(1-\frac{1}{p^s})$
So that we have: $(\phi_s*1)(n)=n^s$
Now set $f=\phi_4$ in the aforementioned equality and we get that:
$\sum_{d=1}^\infty \phi_4(d)1_{d\mid x_1}1_{d\mid x_2}1_{d\mid x_3}1_{d\mid x_4}=\gcd(x_1,x_2,x_3,x_4)^4$
Now we note that:
$\sum_{x_i=1}^\infty\frac{\mu(x_i)}{x_i^2}1_{d\mid x_i}=\sum_{n=1}^\infty\frac{\mu(dn)}{(dn)^2}=\frac{6}{\pi^2}\frac{\mu(d)}{\phi_2(d)}$
Then multiplying both sides of the previous series by $\frac{\mu(x_1)}{x_1^2}\frac{\mu(x_2)}{x_2^2}\frac{\mu(x_3)}{x_3^2}\frac{\mu(x_4)}{x_4^2}$ and rearranging gives:
$\sum_{d=1}^\infty \phi_4(d)(\frac{\mu(x_1)}{x_1^2}1_{d\mid x_1})(\frac{\mu(x_2)}{x_2^2}1_{d\mid x_2})(\frac{\mu(x_3)}{x_3^2}1_{d\mid x_3})(\frac{\mu(x_4)}{x_4^2}1_{d\mid x_4})$ $=\frac{\mu(x_1)\mu(x_2)\mu(x_3)\mu(x_4)}{x_1^2x_2^2x_3^2x_4^2}\gcd(x_1,x_2,x_3,x_4)^4$
Thus:
$\sum_{x_1=1}^\infty\sum_{x_2=1}^\infty\sum_{x_3=1}^\infty\sum_{x_4=1}^\infty\frac{\mu(x_1)\mu(x_2)\mu(x_3)\mu(x_4)}{x_1^2x_2^2x_3^2x_4^2}\gcd(x_1,x_2,x_3,x_4)^4$
$=(\frac{6}{\pi^2})^4\sum_{d=1}^\infty \phi_4(d)\frac{\mu(d)^4}{\phi_2(d)^4}=\frac{1296}{\pi^8}\sum_{n=1}^\infty\frac{\phi_4(n)}{\phi_2(n)^4}|\mu(n)|=\frac{1296}{\pi^8}\prod_{p}(1+\frac{\phi_4(p)}{\phi_2(p)^4})$
So we have:
$\sum_{a=1}^\infty\sum_{b=1}^\infty\sum_{c=1}^\infty\sum_{d=1}^\infty\frac{\mu(a)\mu(b)\mu(c)\mu(d)}{a^2b^2c^2d^2}\gcd(a,b,c,d)^4=\frac{1296}{\pi^8}\prod_{p}(1+\frac{p^4-1}{(p^2-1)^4})$
A similar argument will give your second sum as $\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{\mu(a)\mu(b)}{a^2b^2}\gcd(a,b)^2=\frac{6}{\pi^2}$.
In addition to formula for other such similar generalizations.