Let's solve the problem step by step. We are given a block of mass $5 \mathrm{~kg}$ subjected to a force $F = (-20x + 10) \mathrm{N}$. To find the position and momentum of the block at $t = \frac{\pi}{4}$ seconds, we'll use the equations of motion and integrate the force to find the acceleration, velocity, and position.

First, we start with Newton's second law:

$$F = m \cdot a$$

Given:

$$m = 5 \mathrm{~kg}$$ $$F = -20x + 10$$

So, the acceleration $a$ can be expressed as:

$$a = \frac{F}{m} = \frac{-20x + 10}{5} = -4x + 2$$

Acceleration is the second derivative of position with respect to time:

$$a = \frac{d^2 x}{dt^2} = -4x + 2$$

To solve this differential equation, let's solve it step by step. Assume a solution of the form $x(t) = A \cos(\omega t) + B \sin(\omega t) + C$. Consider using the method of undetermined coefficients:

We first need to find the homogeneous solution (no forcing term):

$$\frac{d^2 x_h}{dt^2} + 4x_h = 0$$

This is a second-order homogeneous differential equation with constant coefficients. Its characteristic equation is:

$$r^2 + 4 = 0$$

Solving for $r$:

$$r = \pm 2i$$

Thus, the general solution to the homogeneous equation is:

$$x_h(t) = A \cos(2t) + B \sin(2t)$$

Next, we need a particular solution to the non-homogeneous equation:

$$x_p = C$$

Substituting $x_p$ into the differential equation:

$$a = -4C + 2 = 0 \implies C = \frac{1}{2}$$

Therefore, the general solution of the differential equation is:

$$x(t) = A \cos(2t) + B \sin(2t) + \frac{1}{2}$$

Using the initial conditions to determine $A$ and $B$:

At $t = 0$, $x(0) = 1$:

$$1 = A \cdot \cos(0) + B \cdot \sin(0) + \frac{1}{2} \implies 1 = A + \frac{1}{2} \implies A = \frac{1}{2}$$

Since the block is initially at rest, $\frac{dx}{dt}|_{t=0} = 0$:

$$\frac{dx}{dt} = -2A \sin(2t) + 2B \cos(2t)$$

At $t = 0$:

$$0 = -2A \cdot \sin(0) + 2B \cdot \cos(0) = 2B \implies B = 0$$

Therefore, the position equation simplifies to:

$$x(t) = \frac{1}{2} \cos(2t) + \frac{1}{2}$$

Now, we find the position at $t = \frac{\pi}{4}$ seconds:

$$x\left(\frac{\pi}{4}\right) = \frac{1}{2} \cos\left(2 \cdot \frac{\pi}{4}\right) + \frac{1}{2} = \frac{1}{2} \cos\left(\frac{\pi}{2}\right) + \frac{1}{2} = \frac{1}{2} \cdot 0 + \frac{1}{2} = \frac{1}{2} \mathrm{~m}$$

To find the momentum, we first find the velocity by differentiating the position function:

$$v(t) = \frac{dx}{dt} = -\sin(2t)$$

Thus, the velocity at $t = \frac{\pi}{4}$ is:

$$v\left(\frac{\pi}{4}\right) = -\sin\left(2 \cdot \frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{2}\right) = -1$$

Finally, the momentum $p$ is given by:

$$p = m \cdot v = 5 \mathrm{~kg} \cdot -1 \mathrm{~m/s} = -5 \mathrm{~kg \cdot m/s}$$

Therefore, the position and momentum of the block at $t = \frac{\pi}{4} \mathrm{s}$ are:

Position: $0.5 \mathrm{~m}$

Momentum: $-5 \mathrm{~kg \cdot m/s}$

The correct option is:

Option C: $0.5 \mathrm{~m}, -5 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}$