Let $P(A\cap B)=0.1=\frac{1}{10}.$

Given that $P(A\mid B)$ and $P(B\mid A)$ are the roots of $12x^2-7x+1=0.$

Now, $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.1}{P(B)},$ $P(B\mid A)=\frac{P(A\cap B)}{P(A)}=\frac{0.1}{P(A)}.$

For the quadratic $12x^2-7x+1=0,$ sum of roots is $\frac{7}{12},$ and product of roots is $\frac{1}{12}.$

Let the roots be $r_1=P(A\mid B),\qquad r_2=P(B\mid A).$

Using the product of roots: $r_1r_2=\frac{1}{12}.$

But $r_1r_2=\frac{0.1}{P(B)}\cdot \frac{0.1}{P(A)}=\frac{0.01}{P(A)P(B)}.$

So, $\frac{0.01}{P(A)P(B)}=\frac{1}{12}$ $P(A)P(B)=0.12.$

Using the sum of roots: $r_1+r_2=\frac{7}{12}.$

That is, $\frac{0.1}{P(B)}+\frac{0.1}{P(A)}=\frac{7}{12}.$

Multiply by $10$: $\frac{1}{P(B)}+\frac{1}{P(A)}=\frac{35}{6}.$

Hence, $\frac{P(A)+P(B)}{P(A)P(B)}=\frac{35}{6}.$

Since $P(A)P(B)=0.12=\frac{3}{25}$, $P(A)+P(B)=\frac{35}{6}\cdot \frac{3}{25}=\frac{7}{10}=0.7.$

Now, $P(\bar A\cup \bar B)=1-P(A\cap B)=1-0.1=0.9.$

Also, $P(\bar A\cap \bar B)=1-P(A\cup B).$

And $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.7-0.1=0.6.$

Therefore, $P(\bar A\cap \bar B)=1-0.6=0.4.$

Required ratio: $\frac{P(\bar A\cup \bar B)}{P(\bar A\cap \bar B)}=\frac{0.9}{0.4}=\frac{9}{4}.$

Hence the correct option is $\boxed{\frac{9}{4}}.$