Answer:
1. 10.7 years
2. 17.0 years
3. 2nd option
Step-by-step explanation:
Use formula for compounded interest

where
A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.
In your case,
1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then



3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

The best will be 2nd option, because $8032.58>$7936.39