1. Use the definitions of conditional probability

Given: $P(E_1\mid E_2)=\frac{1}{2}, \quad P(E_2\mid E_1)=\frac{3}{4}, \quad P(E_1\cap E_2)=\frac{1}{8}$

Now, $P(E_1\mid E_2)=\frac{P(E_1\cap E_2)}{P(E_2)}$ So, $\frac{1}{2}=\frac{1/8}{P(E_2)}$ $P(E_2)=\frac{1}{4}$

Also, $P(E_2\mid E_1)=\frac{P(E_1\cap E_2)}{P(E_1)}$ So, $\frac{3}{4}=\frac{1/8}{P(E_1)}$ $P(E_1)=\frac{1}{6}$

2. Find the remaining basic probabilities

$P(E_1\cap E_2')=P(E_1)-P(E_1\cap E_2)=\frac{1}{6}-\frac{1}{8}=\frac{1}{24}$

$P(E_1'\cap E_2)=P(E_2)-P(E_1\cap E_2)=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$

$P(E_1'\cap E_2')=1-P(E_1\cup E_2)$ with $P(E_1\cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2)=\frac{1}{6}+\frac{1}{4}-\frac{1}{8}=\frac{7}{24}$ Hence, $P(E_1'\cap E_2')=1-\frac{7}{24}=\frac{17}{24}$

Also, $P(E_1')=1-\frac{1}{6}=\frac{5}{6}, \qquad P(E_2')=1-\frac{1}{4}=\frac{3}{4}$

3. Check each option

Option A

$P(E_1\cap E_2)=\frac{1}{8}$ while $P(E_1)P(E_2)=\frac{1}{6}\cdot\frac{1}{4}=\frac{1}{24}$ Since $\frac{1}{8}\ne\frac{1}{24}$ A is false.

Option B

$P(E_1'\cap E_2')=\frac{17}{24}$ while $P(E_1')P(E_2)=\frac{5}{6}\cdot\frac{1}{4}=\frac{5}{24}$ Since $\frac{17}{24}\ne\frac{5}{24}$ B is false.

Option C

$P(E_1\cap E_2')=\frac{1}{24}$ and $P(E_1)P(E_2)=\frac{1}{6}\cdot\frac{1}{4}=\frac{1}{24}$ Thus, $P(E_1\cap E_2')=P(E_1)P(E_2)$ So C is true.

Option D

$P(E_1'\cap E_2)=\frac{1}{8}$ while $P(E_1)P(E_2)=\frac{1}{24}$ Since $\frac{1}{8}\ne\frac{1}{24}$ D is false.

4. Conclusion

The only correct option is: $\boxed{\text{C}}$

This matches the stored correct answer.