I will now go through the questions listed in the image one by one. Let’s begin with the first question:

1. State the cosine and sine laws which are used to solve practical problems involving triangles and angles.
   • Cosine Law: In any triangle, the cosine of an angle is related to the lengths of the sides of the triangle by the formula:
     $c^2 = a^2 + b^2 - 2ab \cdot \cos(C)$ where $a$, $b$, and $c$ are the lengths of the sides of the triangle, and $C$ is the angle opposite side $c$.
   • Sine Law: The sine law relates the sides of a triangle to the sines of its angles: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$ where $a$, $b$, and $c$ are the sides, and $A$, $B$, and $C$ are the respective opposite angles.

2. In a geometric progression, insert 4 geometric terms that are between 2 and 6250.
   • Let the terms be $2, a_2, a_3, a_4, 6250$.
     We know that in a geometric progression, the ratio between consecutive terms is constant. Let $r$ be the common ratio. Then: $2 \cdot r^3 = 6250 \quad \Rightarrow \quad r^3 = \frac{6250}{2} = 3125$ Taking the cube root of both sides: $r = 15$ Therefore, the geometric terms are: $2, 2 \cdot 15 = 30, 30 \cdot 15 = 450, 450 \cdot 15 = 6750, 6250$ Hence, the terms are: $2, 30, 450, 6750$.

3. Solve $e^{x} - 18e^{x} - 3 = 0$.
   • The equation is $e^{x} - 18e^{x} - 3 = 0$.
     First, collect like terms: $-17e^{x} = 3 \quad \Rightarrow \quad e^{x} = -\frac{3}{17}$ However, this leads to a contradiction, since the exponential function $e^x$ cannot be negative for real $x$.
     Therefore, there are no real solutions.

4. Find the differential equation of all straight lines passing through the origin.
   • The general equation of a straight line passing through the origin is: $y = mx$ where $m$ is the slope. Differentiating this with respect to $x$: $\frac{dy}{dx} = m$ Therefore, the differential equation is: $\frac{dy}{dx} = m$ where $m$ is a constant. Since the line passes through the origin, $m$ can vary.

5. Find the vector, parametric, and symmetric equations of the line $l$ passing through the point $A(3, -2, 4)$ with direction vector $\mathbf{u} = (2, 3, 5)$.
   • The vector equation of the line is: $\mathbf{r}(t) = \mathbf{A} + t\mathbf{u} = (3, -2, 4) + t(2, 3, 5)$ which gives the parametric equations: $x = 3 + 2t, \quad y = -2 + 3t, \quad z = 4 + 5t$ The symmetric equations are: $\frac{x - 3}{2} = \frac{y + 2}{3} = \frac{z - 4}{5}$

6. Find square roots of the complex number $3 + 4i$.
   To find the square roots of $3 + 4i$, we express it in polar form: $r = \sqrt{3^2 + 4^2} = 5, \quad \theta = \tan^{-1}\left(\frac{4}{3}\right)$ The square roots of a complex number in polar form $r \text{cis} \theta$ are given by: $\sqrt{r} \text{cis} \left(\frac{\theta + 2k\pi}{2}\right), \quad k = 0, 1$ So the square roots of $3 + 4i$ are: $\pm \left( \sqrt{5} \text{cis} \left(\frac{\theta}{2}\right) \right)$

Would you like more details on these steps, or any further questions?