A motor car costing R200 000 depreciated at a rate of 8% per annum on the reducing balance method. Calculate how long it took for the car to depreciate to a value of R90 000 under these conditions.

$$90\,000 = 200\,000(1 - 0.08)^n$$ $$90\,000 = 200\,000(0.92)^n$$

$$\frac{90\,000}{200\,000} = (0.92)^n$$ $$0.45 = (0.92)^n$$

$$\log(0.45) = n \log(0.92)$$ $$n = \frac{\log(0.45)}{\log(0.92)}$$

$$n \approx \frac{-0.34678}{-0.03621}$$ $$n \approx 9.576$$

It took approximately $\boxed{9.58 \text{ years}}$ for the car to depreciate to R90 000.

3.2.1 To calculate the monthly interest rate:
The annual interest rate is 6.25% compounded semi-annually.
Nominal annual rate $i_{nom} = 0.0625$.
Compounding periods per year $m = 2$.

$$(1 + i_{eff}) = \left(1 + \frac{i_{nom}}{m}\right)^m$$ $$(1 + i_{eff}) = \left(1 + \frac{0.0625}{2}\right)^2$$ $$(1 + i_{eff}) = (1 + 0.03125)^2$$ $$(1 + i_{eff}) = (1.03125)^2 = 1.0634765625$$

$$(1 + i_{monthly})^{12} = (1 + i_{eff})$$ $$(1 + i_{monthly})^{12} = 1.0634765625$$ $$1 + i_{monthly} = (1.0634765625)^{1/12}$$ $$1 + i_{monthly} \approx 1.005139908$$

$$i_{monthly} \approx 1.005139908 - 1$$ $$i_{monthly} \approx 0.005139908$$

The monthly interest rate is approximately $\boxed{0.00514}$ or $0.514\%$.

3.2.2 To calculate the future value of his investment at the end of the five-year period:
The total period is 5 years $\times$ 12 months/year = 60 months.
Monthly interest rate $i = 0.005139908$.
Phetolo makes an initial deposit of R2000 at the beginning of the first month, and then R2000 at the end of the first month, and continues with R2000 at the end of each month for 60 months.

This deposit earns interest for the full 60 months.

$$FV_{\text{initial}} = P(1+i)^n$$ $$FV_{\text{initial}} = 2000(1 + 0.005139908)^{60}$$ $$FV_{\text{initial}} = 2000(1.005139908)^{60}$$ $$FV_{\text{initial}} \approx 2000 \times 1.35620901 \approx \text{R}2712.42$$

$$FV_{\text{annuity}} = X \frac{(1+i)^n - 1}{i}$$ $$FV_{\text{annuity}} = 2000 \frac{(1.005139908)^{60} - 1}{0.005139908}$$ $$FV_{\text{annuity}} = 2000 \frac{1.35620901 - 1}{0.005139908}$$ $$FV_{\text{annuity}} = 2000 \frac{0.35620901}{0.005139908} \approx 2000 \times 69.39000 \approx \text{R}138780.00$$

$$FV_{\text{total}} = FV_{\text{initial}} + FV_{\text{annuity}}$$ $$FV_{\text{total}} = 2712.42 + 138780.00 \approx \text{R}141492.42$$

The future value of his investment is $\boxed{\text{R}141\,492.42}$.

3.2.3 To calculate the value of his investment if he misses the last two payments:
This means only $60 - 2 = 58$ monthly payments were made for the annuity part. The initial R2000 still accumulates for 60 months.

$$FV_{\text{annuity}} = X \frac{(1+i)^n - 1}{i}$$ $$FV_{\text{annuity}} = 2000 \frac{(1.005139908)^{58} - 1}{0.005139908}$$ $$(1.005139908)^{58} \approx 1.34107001$$ $$FV_{\text{annuity}} = 2000 \frac{1.34107001 - 1}{0.005139908}$$ $$FV_{\text{annuity}} = 2000 \frac{0.34107001}{0.005139908} \approx 2000 \times 66.35400 \approx \text{R}132708.00$$

$$FV_{\text{total}} = FV_{\text{initial}} + FV_{\text{annuity}}$$ $$FV_{\text{total}} = 2712.42 + 132708.00 \approx \text{R}135420.42$$

The value of his investment is $\boxed{\text{R}135\,420.42}$.

3.3.1 To determine the number of payments required to settle the loan:
Loan amount $P = \text{R}300\,000$.
Monthly payment $x = \text{R}5\,000$.
Annual interest rate = 18% compounded monthly.
Monthly interest rate $i = \frac{0.18}{12} = 0.015$.

$$P = x \frac{1 - (1+i)^{-n}}{i}$$ $$300\,000 = 5\,000 \frac{1 - (1+0.015)^{-n}}{0.015}$$

$$\frac{300\,000 \times 0.015}{5\,000} = 1 - (1.015)^{-n}$$ $$\frac{4500}{5000} = 1 - (1.015)^{-n}$$ $$0.9 = 1 - (1.015)^{-n}$$ $$(1.015)^{-n} = 1 - 0.9$$ $$(1.015)^{-n} = 0.1$$

$$-n \log(1.015) = \log(0.1)$$ $$-n = \frac{\log(0.1)}{\log(1.015)}$$

$$-n \approx \frac{-1}{0.00646604}$$ $$-n \approx -154.644$$ $$n \approx 154.644$$

Since a payment cannot be a fraction, Jill will make 154 full payments of R5000 and one final smaller payment. Therefore, the total number of payments required is $\boxed{155}$.

3.3.2 To calculate the balance outstanding after Jill has paid the last R5 000:
This refers to the balance outstanding after 154 full payments of R5000.
We use the formula for the balance outstanding on a loan: $A_n = P(1+i)^n - x \frac{(1+i)^n - 1}{i}$.
Here, $n = 154$ (number of full payments).
$P = \text{R}300\,000$.
$x = \text{R}5\,000$.
$i = 0.015$.

$$FV_{\text{loan}} = 300\,000(1+0.015)^{154}$$ $$FV_{\text{loan}} = 300\,000(1.015)^{154}$$ $$(1.015)^{154} \approx 9.8732000$$ $$FV_{\text{loan}} \approx 300\,000 \times 9.8732000 \approx \text{R}2\,961\,960.00$$

$$FV_{\text{payments}} = 5000 \frac{(1.015)^{154} - 1}{0.015}$$ $$FV_{\text{payments}} = 5000 \frac{9.8732000 - 1}{0.015}$$ $$FV_{\text{payments}} = 5000 \frac{8.8732000}{0.015} \approx 5000 \times 591.5466667 \approx \text{R}2\,957\,733.33$$

$$\text{Balance} = FV_{\text{loan}} - FV_{\text{payments}}$$ $$\text{Balance} = 2\,961\,960.00 - 2\,957\,733.33 \approx \text{R}4226.67$$

The balance outstanding after Jill has paid the last R5 000 is $\boxed{\text{R}4226.67}$.