Because the graph passes through $(4,5)$, substitute $x=4$ and $y=5$ into $y=\frac{a}{x-2}+3$:

So the equation is

Now use the point $(k,4)$:

Therefore, the value of $k$ is $6$.

Factor the quadratic:

So for $x\ne -2$,

Then

for all allowed $x$.

The problem says $f(x)$ is equal to the same linear expression for every allowed value of $x$. But the term $\frac{k}{x-3}$ is rational, not linear, unless it disappears entirely. The only way that happens for every allowed $x$ is if

Then

which is linear. Therefore the statement that must be true is $k=0$.

If the original rate is $r$ gallons per minute, then the original filling time is $\frac{720}{r}$ minutes. After the adjustment, the new rate is $r+6$, so the new filling time is $\frac{720}{r+6}$ minutes.

The time decreases by $4$ minutes, so

Combine the fractions:

Multiply both sides by $r(r+6)$:

Divide by $4$:

So

Factor:

This gives $r=-36$ or $r=30$. A filling rate must be positive, so the original rate is $30$ gallons per minute.

Substitute $t=5$ into the formula:

Since $R(5)=8$, we get

Multiply both sides by $2$:

Then subtract $6$:

so

A point $(x,y)$ lies on the graph of $f$ if $y=f(x)$. Evaluate the function at each given $x$-value. For $x=1$,

so $(1,-4)$ is on the graph. Also, $x=2$ is not in the domain because it makes the denominator $0$, and for $x=5$,

not $\frac{8}{5}$. Therefore, the point on the graph is $(1,-4)$.