Set up the differential equation based on the given information.

The rate of change of temperature $X$ with respect to time $t$ is proportional to $X$. Since it's cooling, the constant of proportionality will be negative.

$$\frac{dX}{dt} = -kX$$

where $k$ is a positive constant.

$$\frac{1}{X}\,dX = -k\,dt$$ $$\int \frac{1}{X}\,dX = \int -k\,dt$$ $$\ln|X| = -kt + C_1$$

Exponentiate both sides:

$$|X| = e^{-kt + C_1} = e^{C_1}e^{-kt}$$

Let $A = \pm e^{C_1}$. Since temperature $X$ is positive, $X = Ae^{-kt}$.

At $t=0$, $X=82^\circ C$.

$$82 = Ae^{-k(0)}$$ $$82 = A \cdot 1$$ $$A = 82$$

Therefore, the equation for the temperature is:

$$\boxed{X = 82e^{-kt}}$$

$$42 = 82e^{-k(10)}$$ $$\frac{42}{82} = e^{-10k}$$ $$\frac{21}{41} = e^{-10k}$$

Take the natural logarithm of both sides:

$$\ln\left(\frac{21}{41}\right) = -10k$$ $$k = -\frac{1}{10}\ln\left(\frac{21}{41}\right) = \frac{1}{10}\ln\left(\frac{41}{21}\right)$$ $$k \approx \frac{1}{10} \ln(1.95238) \approx \frac{1}{10} (0.66890) \approx 0.066890$$

$$37 = 82e^{-kt_1}$$ $$\frac{37}{82} = e^{-kt_1}$$ $$\ln\left(\frac{37}{82}\right) = -kt_1$$ $$t_1 = -\frac{1}{k}\ln\left(\frac{37}{82}\right) = \frac{1}{k}\ln\left(\frac{82}{37}\right)$$

Substitute the value of $k$:

$$t_1 = \frac{1}{\frac{1}{10}\ln\left(\frac{41}{21}\right)}\ln\left(\frac{82}{37}\right) = 10 \frac{\ln\left(\frac{82}{37}\right)}{\ln\left(\frac{41}{21}\right)}$$ $$t_1 \approx 10 \frac{\ln(2.216216)}{\ln(1.952381)} \approx 10 \frac{0.79134}{0.66890} \approx 10 \times 1.1829 \approx 11.829 \text{ minutes}$$

The time elapsed after $t=10$ minutes is $t_1 - 10$.

$$\text{Time longer} = 11.829 - 10 = 1.829 \text{ minutes}$$

Rounding to 2 significant figures:

$$\text{Time longer} = \boxed{1.8 \text{ minutes}}$$

Using $k \approx 0.066890$:

$$X = 82e^{-0.066890 \times 3.2}$$ $$X = 82e^{-0.214048}$$ $$X \approx 82 \times 0.80730$$ $$X \approx 66.1986$$

Rounding to 2 significant figures:

$$X \approx \boxed{66^\circ C}$$

Given $\ln y = xy$.
Differentiate both sides:

$$\frac{d}{dx}(\ln y) = \frac{d}{dx}(xy)$$

Using the chain rule for $\ln y$ and the product rule for $xy$:

$$\frac{1}{y}\frac{dy}{dx} = (1)y + x\frac{dy}{dx}$$ $$\frac{1}{y}\frac{dy}{dx} = y + x\frac{dy}{dx}$$

$$\frac{1}{y}\frac{dy}{dx} - x\frac{dy}{dx} = y$$

Factor out $\frac{dy}{dx}$:

$$\frac{dy}{dx}\left(\frac{1}{y} - x\right) = y$$

Combine terms inside the parenthesis:

$$\frac{dy}{dx}\left(\frac{1 - xy}{y}\right) = y$$

Multiply both sides by $\frac{y}{1 - xy}$:

$$\frac{dy}{dx} = y \cdot \frac{y}{1 - xy}$$ $$\frac{dy}{dx} = \boxed{\frac{y^2}{1 - xy}}$$