1. Given probabilities of elementary outcomes

The sample space is $S=\{E_1,E_2,\dots,E_8\}$ and $P(\{E_n\})=\frac{n}{36},\qquad n=1,2,\dots,8.$

So the probabilities are: $\frac{1}{36},\frac{2}{36},\frac{3}{36},\frac{4}{36},\frac{5}{36},\frac{6}{36},\frac{7}{36},\frac{8}{36}.$

Check total probability: $\frac{1+2+3+4+5+6+7+8}{36}=\frac{36}{36}=1.$ So this is valid.

2. Condition on subset $A$

For any subset $A\subseteq S$, $P(A)=\sum_{E_n\in A}\frac{n}{36}.$ We need $P(A)\ge \frac45.$ Multiply by $36$: $\sum_{E_n\in A} n \ge \frac45\cdot 36=28.8.$ Since the left side is an integer, this means $\sum_{E_n\in A} n \ge 29.$

So we must count subsets of $\{1,2,3,4,5,6,7,8\}$ whose element-sum is at least $29$.

3. Use complement counting

The total sum is $1+2+3+4+5+6+7+8=36.$

Let the complement of $A$ be $A^c$. Then $\sum_{n\in A} n = 36-\sum_{n\in A^c} n.$ Condition $\sum_{n\in A} n\ge 29$ is equivalent to $36-\sum_{n\in A^c} n \ge 29,$ so $\sum_{n\in A^c} n \le 7.$

Thus the required number of subsets $A$ equals the number of subsets of $\{1,2,3,4,5,6,7,8\}$ whose sum is at most $7$.

Since any element bigger than $7$ cannot appear in such a subset, we only need subsets of $\{1,2,3,4,5,6,7\}$ with sum $\le 7$.

4. List all subsets with sum $\le 7$

We count by sum.

• Sum $0$: $\varnothing$ Count = 1

• Sum $1$: $\{1\}$ Count = 1

• Sum $2$: $\{2\}$ Count = 1

• Sum $3$: $\{3\},\{1,2\}$ Count = 2

• Sum $4$: $\{4\},\{1,3\}$ Count = 2

• Sum $5$: $\{5\},\{1,4\},\{2,3\}$ Count = 3

• Sum $6$: $\{6\},\{1,5\},\{2,4\},\{1,2,3\}$ Count = 4

• Sum $7$: $\{7\},\{1,6\},\{2,5\},\{3,4\},\{1,2,4\}$ Count = 5

Total number of such subsets: $1+1+1+2+2+3+4+5=19.$

5. Conclusion

Hence the number of subsets $A\subseteq S$ such that $P(A)\ge \frac45$ is $\boxed{19}.$

6. Comparison with stored answer

Stored correct answer = $19$.

Our derived answer is also $19$, so they agree.