Part 1: Finding the frequency of the sinewave

The frequency $f$ of a sinewave is the reciprocal of the period $T$, which is given by the formula:

$f = \frac{1}{T}$

Here, the period $T$ is 535 μs (microseconds). First, we need to convert this into seconds:

$T = 535 \times 10^{-6} \, \text{seconds}$

Now, we can calculate the frequency:

$f = \frac{1}{535 \times 10^{-6}} \, \text{Hz}$

$f \approx 1870.56 \, \text{Hz}$

To express this in kHz:

$f = \frac{1870.56}{1000} \, \text{kHz} \approx 1.871 \, \text{kHz}$

So, the frequency is approximately 1.871 kHz to three significant figures.

Part 2: Finding the new period after the frequency increase

The sinewave's frequency is now increased by a factor of 4.25. Let's first calculate the new frequency:

$f_{\text{new}} = 1.87056 \times 4.25 \, \text{kHz}$

Now we calculate the new period $T_{\text{new}}$, remembering that the period is the reciprocal of the frequency:

$T_{\text{new}} = \frac{1}{f_{\text{new}}}$

I'll now compute the new frequency and period using the exact values for precision.