Let's solve step by step each problem:
A)
Here is important to use the multiplication property. When the base is the same you can sum the exponents:
![8^7*8^(-12)=8^(7+(-12))=8^(7-12)=8^(-5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/uz45uh2ywrgnf3omzlg150f6u37j23viw6.png)
Now you can see the exponent is negative and you can use another property to turn him positive. This can be done by inverting the fraction:
![8^(-5)=(1)/(8^5)](https://img.qammunity.org/2023/formulas/mathematics/high-school/c8o36kdxcu1qscghuedgqml63xfi6ur1ff.png)
8^5 is such a huge number and definitely, this answer will be between 0 and 1. Just showing you the value:
![(1)/(8^5)=0.00003](https://img.qammunity.org/2023/formulas/mathematics/high-school/bqodwu3fu1bz80ijnouqt79khms5amqmk0.png)
B)
Here you have to use the division property. When the base is the same you can subtract the exponents:
![(7^4)/(7^(-3))=7^(4-(-3))=7^7=823543](https://img.qammunity.org/2023/formulas/mathematics/high-school/wn5xz6zaicn3md89s9leo1j3ee1ssfo123.png)
(not between 0 and 1)
C) Use the multiplication property that we used on letter A)
![((1)/(3))^2*((1)/(3))^9=((1)/(3))^(2+9)=((1)/(3))^(11)](https://img.qammunity.org/2023/formulas/mathematics/high-school/2w5ykyv72xkmw3eabiy4yj2oc7b5jw5tyt.png)
![((1)/(3))^(11)=(1^(11))/(3^(11))=0.000005^{}](https://img.qammunity.org/2023/formulas/mathematics/high-school/4zqle0dmvn1lfkjgas6u4vea3nei3ja2ak.png)
Between 0 and 1!
D) Use the division property that we used on letter B)
![((-5)^6)/((-5)^(10))=(-5)^((6-10))=(-5)^(-4)](https://img.qammunity.org/2023/formulas/mathematics/high-school/lqh9eu5jexj67jjcixex2e3r7uqbtajbc4.png)
![(-5)^(-4)=(1)/((-5)^4)=0.0016](https://img.qammunity.org/2023/formulas/mathematics/high-school/tqm374en0mee96ho5oxpet6nvnlmhz6flb.png)
Between 0 and 1! Therefore:
A, C and D are between 0 and 1.
B is not between 0 and 1.