If you are trying to maximise the expected score, then since the expected value of the 12-sided die is $6.5$, it makes sense to stop when the 8-sided die shows greater than $6.5$, i.e. when it shows $7$ or $8$, each with probability $\frac18$. So with probability $\frac34$ you throw the 12-sided die.

The expected score is then $$7 \times \frac18 + 8 \times \frac18 + 6.5 \times \frac34$$

If you are trying to maximise the expected score, then since the expected value of the 12-sided die is $6.5$, it makes sense to stop when the 8-sided die shows greater than $6.5$, i.e. when it shows $7$ or $8$, each with probability $\frac18$. So with probability $\frac34$ you throw the 12-sided die.

The expected score is then $$7 \times \frac18 + 8 \times \frac18 + 6.5 \times \frac34 = 6.75.$$