A circle has its centre at M(-2; 1) and passes through S(1; -2). Determine the coordinates of M and the radius of the circle. Determine the equation of the tangent RS.

The equation of the circle with centre M is given as $(x+2)^2 + (y-1)^2 = r^2$.
Comparing this to the standard form of a circle equation $(x-a)^2 + (y-b)^2 = r^2$, the coordinates of the centre M are $(-2; 1)$.
The point S$(1; -2)$ lies on the circle. Substitute these coordinates into the equation to find $r^2$:

$$(1+2)^2 + (-2-1)^2 = r^2$$ $$(3)^2 + (-3)^2 = r^2$$ $$9 + 9 = r^2$$ $$r^2 = 18$$

The radius $r = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$.
The coordinates of M are $\boxed{(-2; 1)}$ and the radius is $\boxed{3\sqrt{2}}$.

The tangent RS touches the circle at point S$(1; -2)$. The radius MS is perpendicular to the tangent at S.
First, calculate the gradient of the radius MS, $m_{MS}$, using M$(-2; 1)$ and S$(1; -2)$:

$$m_{MS} = \frac{y_S - y_M}{x_S - x_M} = \frac{-2 - 1}{1 - (-2)} = \frac{-3}{1 + 2} = \frac{-3}{3} = -1$$

Since the tangent RS is perpendicular to the radius MS, the gradient of the tangent, $m_{RS}$, is the negative reciprocal of $m_{MS}$:

$$m_{RS} = -\frac{1}{m_{MS}} = -\frac{1}{-1} = 1$$

Now, use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with point S$(1; -2)$ and $m_{RS} = 1$:

$$y - (-2) = 1(x - 1)$$ $$y + 2 = x - 1$$ $$y = x - 1 - 2$$

The equation of the tangent RS is $\boxed{y = x - 3}$.

The circles with centres P and M touch externally at point S. This means S lies on the line segment MP.
We are given the ratio MS : MP = 1 : 3.
This implies that the distance from M to S is one-third of the distance from M to P.
Let M be $(-2; 1)$ and S be $(1; -2)$. Let P be $(a; b)$.
We can use the section formula or vector approach. Using the vector approach:
The vector $\vec{MS}$ represents the displacement from M to S:

$$\vec{MS} = (x_S - x_M, y_S - y_M) = (1 - (-2), -2 - 1) = (3, -3)$$

Since MS : MP = 1 : 3, the vector $\vec{MP}$ is 3 times the vector $\vec{MS}$:

$$\vec{MP} = 3 \times \vec{MS} = 3(3, -3) = (9, -9)$$

The vector $\vec{MP}$ also represents the displacement from M to P:

$$\vec{MP} = (x_P - x_M, y_P - y_M) = (a - (-2), b - 1) = (a+2, b-1)$$

Equating the components of $\vec{MP}$:

$$a+2 = 9 \implies a = 7$$ $$b-1 = -9 \implies b = -8$$

The coordinates of point P are $\boxed{(7; -8)}$.