1. Given data

We are given: $P(E)=\frac18,\quad P(F)=\frac16,\quad P(G)=\frac14,\quad P(E\cap F\cap G)=\frac1{10}.$

We must test each statement.

2. First check consistency of the given probabilities

For any three events, $E\cap F\cap G \subseteq E,\quad E\cap F\cap G \subseteq F,\quad E\cap F\cap G \subseteq G.$ Hence, $P(E\cap F\cap G) \le P(E),\ P(F),\ P(G).$

But here, $P(E\cap F\cap G)=\frac1{10}=0.1,$ while $P(E)=\frac18=0.125,\quad P(F)=\frac16\approx 0.1667,\quad P(G)=\frac14=0.25.$ So the data is not immediately impossible from this check.

However, also note that $E\cap F\cap G \subseteq E\cap F,$ so $P(E\cap F) \ge \frac1{10},$ and similarly for other pairwise intersections.

We now evaluate each option carefully.

3. Option A: $P(E\cap F\cap G^c) \le \frac1{40}$

Observe that $E\cap F=(E\cap F\cap G)\ \cup\ (E\cap F\cap G^c),$ with disjoint union. Therefore, $P(E\cap F\cap G^c)=P(E\cap F)-P(E\cap F\cap G).$

Now, $P(E\cap F) \le P(E)=\frac18.$ Hence, P(E\cap F\cap G^c) \le \frac18-\frac1{10}=rac{5-4}{40}=\frac1{40}.

So A is true.

4. Option B: $P(E^c\cap F\cap G) \le \frac1{15}$

Similarly, $F\cap G=(E\cap F\cap G)\ \cup\ (E^c\cap F\cap G),$ so $P(E^c\cap F\cap G)=P(F\cap G)-P(E\cap F\cap G).$

Since $P(F\cap G)\le P(F)=\frac16,$ we get P(E^c\cap F\cap G) \le \frac16-\frac1{10}=rac{5-3}{30}=\frac1{15}.

So B is true.

5. Option C: $P(E\cup F\cup G) \le \frac{13}{24}$

Using inclusion-exclusion, $P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(F\cap G)-P(G\cap E)+P(E\cap F\cap G).$

Now, $P(E\cap F)\ge P(E\cap F\cap G)=\frac1{10},$ $P(F\cap G)\ge \frac1{10},$ $P(G\cap E)\ge \frac1{10}.$

Therefore, $P(E\cup F\cup G) \le \frac18+\frac16+\frac14-3\cdot\frac1{10}+\frac1{10}.$

Simplify: $\frac18+\frac16+\frac14=\frac{3+4+6}{24}=\frac{13}{24}.$ Thus, P(E\cup F\cup G) \le \frac{13}{24}-\frac{2}{10}=rac{13}{24}-\frac15. Since subtracting a positive number makes it even smaller, certainly $P(E\cup F\cup G) \le \frac{13}{24}.$

Hence C is true.

6. Option D: $P(E^c\cup F^c\cup G^c) \le \frac{5}{12}$

By De Morgan's law, $E^c\cup F^c\cup G^c=(E\cap F\cap G)^c.$ Therefore, $P(E^c\cup F^c\cup G^c)=1-P(E\cap F\cap G)=1-\frac1{10}=\frac9{10}.$

Now, $\frac9{10}=0.9 > \frac5{12}\approx 0.4167.$ So the inequality is false.

Hence D is false.

7. Final conclusion

The true statements are: $\boxed{A,\ B,\ C}.$