The equation of the hyperbola is given as (x+2)(y+3) = t.

Here are the solutions to Question 1:

1.1.1
The equation of the hyperbola is given as $(x+2)(y+3) = t$.
The vertical asymptote is found by setting the factor involving $x$ to zero:

$$x+2 = 0 \implies x = -2$$

The horizontal asymptote is found by setting the factor involving $y$ to zero:

$$y+3 = 0 \implies y = -3$$

The equations of the asymptotes of $g$ are $\boxed{x = -2 \text{ and } y = -3}$.

1.1.2
1.1.2.1 Determine the value of $t$.
We are given that $g(0) = -\frac{5}{2}$. This means when $x=0$, $y=-\frac{5}{2}$.
Substitute these values into the equation $(x+2)(y+3) = t$:

$$(0+2)\left(-\frac{5}{2}+3\right) = t$$ $$(2)\left(-\frac{5}{2}+\frac{6}{2}\right) = t$$ $$(2)\left(\frac{1}{2}\right) = t$$ $$t = \boxed{1}$$

1.1.2.2 Determine the values of $p$ and $q$.
The function $f(x) = (x+p)^2 + q$ is a parabola. Its axis of symmetry is $x = -p$.
The problem states that one of the asymptotes of $g$ is an axis of symmetry for $f$. From 1.1.1, the asymptotes of $g$ are $x=-2$ and $y=-3$. Since the axis of symmetry for a parabola is a vertical line, it must be $x=-2$.
Therefore, $-p = -2 \implies p = 2$.
The turning point of the parabola $f$ is $(-p, q)$. Since $p=2$, the x-coordinate of the turning point is $-2$. From the graph, the turning point of $f$ is at $(-2, -1)$.
Thus, $q = -1$.
The values are $\boxed{p = 2 \text{ and } q = -1}$.

1.2
Write $g$ in the form $y = \frac{a}{x+p} + q$.
From 1.1.2.1, we found $t=1$. So the equation for $g$ is $(x+2)(y+3) = 1$.
Divide by $(x+2)$:

$$y+3 = \frac{1}{x+2}$$

Subtract 3 from both sides:

$$y = \frac{1}{x+2} - 3$$

So, $g$ in the required form is $\boxed{y = \frac{1}{x+2} - 3}$.

1.3
If $h(x) = x-1$ is the line of symmetry to $g$, determine the coordinates of the points of intersection $h$ and $g$.
To find the points of intersection, we set the equations for $h(x)$ and $g(x)$ equal to each other.
Equation for $h$: $y = x-1$
Equation for $g$: $(x+2)(y+3) = 1$ (using $t=1$ from 1.1.2.1)
Substitute $y = x-1$ into the equation for $g$:

$$(x+2)((x-1)+3) = 1$$ $$(x+2)(x+2) = 1$$ $$(x+2)^2 = 1$$

Take the square root of both sides:

$$x+2 = \pm \sqrt{1}$$ $$x+2 = \pm 1$$

Case 1: $x+2 = 1$

$$x = 1-2$$ $$x = -1$$

Substitute $x=-1$ into $y=x-1$:

$$y = -1-1$$ $$y = -2$$

Point 1: $(-1, -2)$

Case 2: $x+2 = -1$

$$x = -1-2$$ $$x = -3$$

Substitute $x=-3$ into $y=x-1$:

$$y = -3-1$$ $$y = -4$$

Point 2: $(-3, -4)$
The coordinates of the points of intersection are $\boxed{(-1, -2) \text{ and } (-3, -4)}$.

1.4
If $k = x^2+4x+3$. Determine the values of $k$ if its roots are:
We consider the quadratic equation $x^2+4x+3-k=0$.
The discriminant is $\Delta = B^2 - 4AC$. Here, $A=1$, $B=4$, $C=3-k$.

$$\Delta = (4)^2 - 4(1)(3-k)$$ $$\Delta = 16 - 12 + 4k$$ $$\Delta = 4 + 4k$$

1.4.1 non-real roots
For non-real roots, the discriminant must be less than zero ($\Delta < 0$).

$$4 + 4k < 0$$ $$4k < -4$$ $$k < \boxed{-1}$$

1.4.2 negative and unequal roots
For unequal roots, the discriminant must be greater than zero ($\Delta > 0$).

$$4 + 4k > 0$$ $$4k > -4$$ $$k > -1$$

For negative roots, two conditions must be met:

1. The sum of the roots ($x_1+x_2 = -\frac{B}{A}$) must be negative. $$x_1+x_2 = -\frac{4}{1} = -4$$ This condition is satisfied as $-4 < 0$.
2. The product of the roots ($x_1x_2 = \frac{C}{A}$) must be positive. $$x_1x_2 = \frac{3-k}{1} = 3-k$$ $$3-k > 0$$ $$-k > -3$$ $$k < 3$$

Combining the conditions $k > -1$ and $k < 3$, we get:

$$\boxed{-1 < k < 3}$$

1.5
1.5.1 domain of $g$
The function $g(x)$ is a hyperbola with a vertical asymptote at $x=-2$. The function is undefined at this value.
The domain of $g$ is $\boxed{x \in \mathbb{R}, x \neq -2}$.

1.5.2 range of $f$
The function $f(x) = (x+p)^2 + q$ is a parabola. From 1.1.2.2, we found $p=2$ and $q=-1$.
So, $f(x) = (x+2)^2 - 1$.
This is a parabola that opens upwards (since the coefficient of $(x+2)^2$ is positive). Its minimum value occurs at the turning point $(-2, -1)$.
The minimum y-value is $-1$.
The range of $f$ is $\boxed{y \ge -1}$.