Determine the equation of the circle with diameter NP, where N(-2; 3) and P(4; -7). Then, find the equation of the tangent to the circle at point N.

The diameter of the circle is NP, with N$(-2; 3)$ and P$(4; -7)$.
The center of the circle $(a; b)$ is the midpoint of the diameter NP.

$$a = \frac{x_N + x_P}{2} = \frac{-2 + 4}{2} = \frac{2}{2} = 1$$ $$b = \frac{y_N + y_P}{2} = \frac{3 + (-7)}{2} = \frac{-4}{2} = -2$$

The center of the circle is $(1; -2)$.

The radius squared, $r^2$, can be found using the distance from the center $(1; -2)$ to N$(-2; 3)$.

$$r^2 = (x_N - a)^2 + (y_N - b)^2$$ $$r^2 = (-2 - 1)^2 + (3 - (-2))^2$$ $$r^2 = (-3)^2 + (5)^2$$ $$r^2 = 9 + 25$$ $$r^2 = 34$$

Using the center $(a; b) = (1; -2)$ and $r^2 = 34$, the equation of the circle is:

$$(x - a)^2 + (y - b)^2 = r^2$$ $$(x - 1)^2 + (y - (-2))^2 = 34$$ $$\boxed{(x - 1)^2 + (y + 2)^2 = 34}$$

The tangent to the circle at N$(-2; 3)$ is perpendicular to the radius CN, where C is the center $(1; -2)$.
First, calculate the gradient of the radius CN:

$$m_{CN} = \frac{y_N - y_C}{x_N - x_C} = \frac{3 - (-2)}{-2 - 1} = \frac{5}{-3} = -\frac{5}{3}$$

The gradient of the tangent line, $m_{tangent}$, is the negative reciprocal of $m_{CN}$:

$$m_{tangent} = -\frac{1}{m_{CN}} = -\frac{1}{-\frac{5}{3}} = \frac{3}{5}$$

Now, use the point-slope form $y - y_1 = m(x - x_1)$ with N$(-2; 3)$ and $m_{tangent} = \frac{3}{5}$:

$$y - 3 = \frac{3}{5}(x - (-2))$$ $$y - 3 = \frac{3}{5}(x + 2)$$ $$y - 3 = \frac{3}{5}x + \frac{6}{5}$$ $$y = \frac{3}{5}x + \frac{6}{5} + 3$$ $$y = \frac{3}{5}x + \frac{6}{5} + \frac{15}{5}$$ $$\boxed{y = \frac{3}{5}x + \frac{21}{5}}$$

Given N$(-2; 3)$, P$(4; -7)$, and Q$(4; 3)$. NP is the diameter.
An angle subtended by a diameter at any point on the circumference is $90^\circ$. Therefore, $N\hat{Q}P = 90^\circ$.
This means $\triangle NPQ$ is a right-angled triangle at Q.
Calculate the lengths of the sides NQ and PQ:

$$NQ = \sqrt{(x_Q - x_N)^2 + (y_Q - y_N)^2} = \sqrt{(4 - (-2))^2 + (3 - 3)^2} = \sqrt{(6)^2 + (0)^2} = \sqrt{36} = 6$$ $$PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} = \sqrt{(4 - 4)^2 + (3 - (-7))^2} = \sqrt{(0)^2 + (10)^2} = \sqrt{100} = 10$$

In the right-angled triangle $\triangle NPQ$, we can use trigonometry to find $N\hat{P}Q$:

$$\tan(N\hat{P}Q) = \frac{\text{opposite}}{\text{adjacent}} = \frac{NQ}{PQ} = \frac{6}{10} = \frac{3}{5}$$ $$N\hat{P}Q = \arctan\left(\frac{3}{5}\right)$$ $$N\hat{P}Q \approx 30.9637^\circ$$

Rounding to one decimal place:

$$\boxed{N\hat{P}Q \approx 31.0^\circ}$$

First, calculate the area of $\triangle PQN$. Since it's a right-angled triangle at Q:

$$\text{Area of } \triangle PQN = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times NQ$$