## Of course, I'll be happy to help you with your questions!

#1. To find the intensity outside the soundproofed room, we can use the equation:

I/Io = 10^(dB/10)

In this case, we know that the soundproofed room is 49.7 dB quieter than the outside. The sound intensity in the room is given as 2.66 x 10^-10 W/m^2.

Plugging in these values, we have:

I/Io = 10^(49.7/10)

Calculating this, we find that I/Io is approximately 9.33 x 10^4.97.

Now, to find the intensity outside (I), we multiply the known intensity inside the room (2.66 x 10^-10 W/m^2) by the value of I/Io:

I = (2.66 x 10^-10) * 9.33 x 10^4.97

Evaluating this calculation, we obtain the intensity outside the soundproofed room.

#2. To determine the change in sound intensity level when the listener increases their distance from the sound source by a factor of 3.86, we'll use the fact that sound intensity falls off with the square of the distance.

The change in intensity level is given by the equation:

Change in intensity level = -10 * log10(1/factor^2)

In this case, the factor is 3.86. Plugging this into the equation, we have:

Change in intensity level = -10 * log10(1/3.86^2)

Evaluating this expression, we find that the change in intensity level is approximately 5.86 dB less intense.

I hope this helps! Let me know if you have any further questions.