Given that cos p = 7/25, 0° < p < 90°, find the value of sin p.

A rational expression is undefined when its denominator is equal to zero.
The given expression is $\frac{m+1}{3m^2 - 2m - 5}$.
Set the denominator to zero:

$$3m^2 - 2m - 5 = 0$$

We can factor the quadratic equation. We need two numbers that multiply to $3 \times -5 = -15$ and add to $-2$. These numbers are $-5$ and 3.
Rewrite the middle term:

$$3m^2 - 5m + 3m - 5 = 0$$

Factor by grouping:

$$m(3m - 5) + 1(3m - 5) = 0$$ $$(3m - 5)(m + 1) = 0$$

Set each factor to zero to find the values of $m$:

$$3m - 5 = 0 \Rightarrow 3m = 5 \Rightarrow m = \frac{5}{3}$$ $$m + 1 = 0 \Rightarrow m = -1$$

The expression is not defined for $m = -1$ and $m = \frac{5}{3}$.

The correct option is D.

$$\boxed{\text{D. } -1 \text{ and } \frac{5}{3}}$$

2. Finding the value of $\sin p$

Given $\cos p = \frac{7}{25}$ and $0^\circ < p < 90^\circ$.
Substitute the value of $\cos p$ into the identity:

$$\sin^2 p + \left(\frac{7}{25}\right)^2 = 1$$ $$\sin^2 p + \frac{49}{625} = 1$$

Subtract $\frac{49}{625}$ from both sides:

$$\sin^2 p = 1 - \frac{49}{625}$$ $$\sin^2 p = \frac{625}{625} - \frac{49}{625}$$ $$\sin^2 p = \frac{625 - 49}{625}$$ $$\sin^2 p = \frac{576}{625}$$

Take the square root of both sides:

$$\sin p = \sqrt{\frac{576}{625}}$$

Since $0^\circ < p < 90^\circ$, $p$ is in the first quadrant, so $\sin p$ must be positive.

$$\sin p = \frac{\sqrt{576}}{\sqrt{625}}$$ $$\sin p = \frac{24}{25}$$

The correct option is B.

$$\boxed{\text{B. } \frac{24}{25}}$$