Let $$E$$ & $$F$$ be two independent events. The probability that $$E$$ & $$F$$ happen is $$\frac{1}{12}$$ and the probability that neither $$E$$ nor $$F$$ happens is $$\frac{1}{2}$$, then a value of $$\frac{P(E)}{P(F)}$$ is:

Let the probabilities of the two independent events be denoted by

$$P(E)=p,\;P(F)=q\qquad -(1)$$

Because the events are independent, the probability of their simultaneous occurrence is simply the product of their individual probabilities. The question tells us that this probability equals $$\frac{1}{12}$$, so

$$pq=\frac{1}{12}\qquad -(2)$$

It is also given that neither event happens with probability $$\frac{1}{2}$$. For independent events, the probability that both fail is the product of their individual failure probabilities:

$$P(E'\cap F')=(1-p)(1-q)=\frac{1}{2}\qquad -(3)$$

Now expand the left‐hand side of $$(3)$$:

$$(1-p)(1-q)=1-p-q+pq\qquad -(4)$$

Substituting $$pq$$ from $$(2)$$ into $$(4)$$ gives

$$1-p-q+\frac{1}{12}=\frac{1}{2}\qquad -(5)$$

Move all constant terms to one side and the variables to the other:

$$1+\frac{1}{12}-\frac{1}{2}=p+q\qquad -(6)$$

Convert everything to a common denominator $$12$$ inside $$(6)$$:

$$\frac{12}{12}+\frac{1}{12}-\frac{6}{12}=p+q\qquad -(7)$$

Simplifying $$(7)$$ gives

$$\frac{7}{12}=p+q\qquad -(8)$$

So we now have two simultaneous relations, $$(2)$$ and $$(8)$$:

$$pq=\frac{1}{12},\qquad p+q=\frac{7}{12}\qquad -(9)$$

Solve for one variable in terms of the other using $$(8)$$; let us write

$$q=\frac{7}{12}-p\qquad -(10)$$

Substitute $$(10)$$ into $$(2)$$:

$$p\left(\frac{7}{12}-p\right)=\frac{1}{12}\qquad -(11)$$

Multiply both sides of $$(11)$$ by $$12$$ to remove denominators:

$$7p-12p^{2}=1\qquad -(12)$$

Rearrange $$(12)$$ into the standard quadratic form:

$$12p^{2}-7p+1=0\qquad -(13)$$

Compute the discriminant $$D$$ of $$(13)$$:

$$D=(-7)^{2}-4(12)(1)=49-48=1\qquad -(14)$$

The square root of the discriminant is $$\sqrt{D}=1$$, so the quadratic formula yields

$$p=\frac{7\pm1}{2\cdot12}=\frac{7\pm1}{24}\qquad -(15)$$

This gives two possible values:

$$p=\frac{8}{24}=\frac{1}{3}\quad\text{or}\quad p=\frac{6}{24}=\frac{1}{4}\qquad -(16)$$

Using $$(10)$$ to find the corresponding $$q$$ values:

If $$p=\frac{1}{3}$$, then $$q=\frac{7}{12}-\frac{1}{3}=\frac{7}{12}-\frac{4}{12}=\frac{3}{12}=\frac{1}{4}\qquad -(17)$$

If $$p=\frac{1}{4}$$, then $$q=\frac{7}{12}-\frac{1}{4}=\frac{7}{12}-\frac{3}{12}=\frac{4}{12}=\frac{1}{3}\qquad -(18)$$

Thus the unordered pair $$\{p,q\}$$ is $$\left\{\frac{1}{3},\frac{1}{4}\right\}$$, and either event could take either value. The required ratio is

$$\frac{P(E)}{P(F)}=\frac{p}{q}\qquad -(19)$$

When $$p=\frac{1}{3},\;q=\frac{1}{4}$$, equation $$(19)$$ gives

$$\frac{1/3}{1/4}=\frac{1}{3}\times4=\frac{4}{3}\qquad -(20)$$

If the roles are reversed, $$(19)$$ would give $$\frac{1}{4}/\frac{1}{3}=\frac{3}{4}$$, which is not among the answer choices. Hence the admissible value that matches an option is

$$\frac{P(E)}{P(F)}=\frac{4}{3}\qquad -(21)$$

Hence, the correct answer is Option A.