Consider the functions g(x) = a^x and its reflection in the y-axis, h(x). Point A is (-3; 27/8) and point B is the y-intercept of g(x). Find the coordinates of B, the value of a, the equation of h(x),

The y-intercept occurs when $x=0$.

The coordinates of B are $\boxed{\mathbf{(0; 1)}}$.

Substitute the coordinates of A into the equation:

The value of $a$ is $\boxed{\mathbf{\frac{2}{3}}}$.

A reflection in the y-axis means replacing $x$ with $-x$ in the function's equation.

The equation of $h$ is $\boxed{\mathbf{h(x) = \left(\frac{3}{2}\right)^x}}$.

The equation of $h^{-1}(x)$ is $\boxed{\mathbf{y = \log_{\frac{3}{2}} x}}$.

Since the base $\frac{3}{2} > 1$, the inequality direction remains the same.

For $h^{-1}(x) = \log_{\frac{3}{2}} x$, the domain requires $x > 0$.
Combining the conditions, $0 < x \le \frac{3}{2}$.
The values of $x$ are $\boxed{\mathbf{0 < x \le \frac{3}{2}}}$.

The domain of $h^{-1}(x) = \log_{\frac{3}{2}} x$ is $x > 0$.

For $h^{-1}(x-3)$, the expression inside the logarithm, $(x-3)$, must be greater than 0.

The domain of $h^{-1}(x-3)$ is $\boxed{\mathbf{x > 3}}$.

For a hyperbola $f(x) = \frac{a}{x+p} + q$, the horizontal asymptote is $y=q$. The range is $y \in \mathbb{R}, y \ne q$.

The range of $f$ is $y \in \mathbb{R}, y \ne 2$.
Therefore, $q=2$.
The value of $q$ is $\boxed{\mathbf{2}}$.

From 6.1, $q=2$, so $f(x) = \frac{a}{x+p} + 2$.

The vertical asymptote is $x = -p$. The point of intersection of the asymptotes is $(-p; q)$, which is $(-p; 2)$.
The axes of symmetry pass through this point. The given axis of symmetry is $y = x+1$.
Substitute $(-p; 2)$ into $y = x+1$:

Now the function is $f(x) = \frac{a}{x-1} + 2$.
The graph passes through $(0; -4)$. Substitute this point into the function:

The values are $\boxed{\mathbf{a=6 \text{ and } p=-1}}$.

From $f(x) = \frac{6}{x-1} + 2$:
Vertical asymptote: $x=1$.
Horizontal asymptote: $y=2$.

Y-intercept (set $x=0$):

The y-intercept is $(0; -4)$.
X-intercept (set $f(x)=0$):

The x-intercept is $(-2; 0)$.

Plot the vertical asymptote $x=1$ and the horizontal asymptote $y=2$.
Plot the intercepts $(0; -4)$ and $(-2; 0)$.
Draw the two branches of the hyperbola, approaching the asymptotes and passing through the intercepts.

graph TD
    A[Start] --> B{Plot Asymptotes};
    B --> C[Vertical: x=1];
    B --> D[Horizontal: y=2];
    B --> E{Find Intercepts};
    E --> F[Y-intercept: (0, -4)];
    E --> G[X-intercept: (-2, 0)];
    E --> H{Sketch Hyperbola Branches};
    H --> I[Pass through intercepts];
    H --> J[Approach asymptotes];
    J --> K[End];

style A fill:#fff,stroke:#333,stroke-width:2px
style B fill:#fff,stroke:#333,stroke-width:2px
style C fill:#fff,stroke:#333,stroke-width:2px
style D fill:#fff,stroke:#333,stroke-width:2px
style E fill:#fff,stroke:#333,stroke-width:2px
style F fill:#fff,stroke:#333,stroke-width:2px
style G fill:#fff,stroke:#333,stroke-width:2px
style H fill:#fff,stroke:#333,stroke-width:2px
style I fill:#fff,stroke:#333,stroke-width:2px
style J fill:#fff,stroke:#333,stroke-width:2px
style K fill:#fff,stroke:#333,stroke-width:2px

The graph should look like this:
(Due to text-based format, I cannot draw the graph directly. Please sketch it on graph paper using the information above.)