A function g has equation (x+2)(y+3)=t. The function f(x)=(x+p)2+q has an axis of symmetry which is one of the asymptotes of g. If g(0) = -5/2, determine the values of t, p, q.

For $g(x): (x+2)(y+3) = t$:
The vertical asymptote is $x+2=0 \implies x=-2$.
The horizontal asymptote is $y+3=0 \implies y=-3$.
The equations of the asymptotes are $\boxed{x=-2 \text{ and } y=-3}$.

1.1.2 Determine the values of:

1.1.2.1 $t$
We are given that $g(0) = -\frac{5}{2}$. This means the point $(0, -\frac{5}{2})$ lies on the graph of $g$.

1.1.2.2 $p$ and $q$
The function $f(x) = (x+p)^2 + q$ is a parabola. The problem states that one of the asymptotes of $g$ is an axis of symmetry for $f$.

The axis of symmetry for $f(x) = (x+p)^2 + q$ is the vertical line $x = -p$.
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 vertex of the parabola $f(x) = (x+p)^2 + q$ is at $(-p, q)$.
With $p=2$, the vertex is at $(-2, q)$.
From the sketch, the parabola $f$ touches the x-axis at $x=-2$, which means its vertex is at $(-2, 0)$.
Therefore, $q=0$.
The values are $p=\boxed{2}$ and $q=\boxed{0}$.