We are given a set of questions about functions and continuity. We need to find the answers to these questions based on the given information.

1. Problem Description

We are given a set of questions about functions and continuity. We need to find the answers to these questions based on the given information.

2. Solution Steps

Let's address the questions one by one based on the given (partially visible) context. Because the image is low quality, many of the expressions are hard to determine and my answers are only based on educated guesses based on the context.

Question 6: "If $d(s) = s$ ... $\frac{... - 1}{...}$ ... $\tan s$, $d(0) = p$. Then the value of $p$ that makes the function ... is ..."

Without being able to read more of the expression for $d(s)$, it is hard to say. However, since we are told that $d(0) = p$, we need to find $d(0)$. If we assume that $d(s)$ is merely $d(s) = \tan s$, then $d(0) = \tan 0 = 0$, thus $p=0$. However, it may be that we must consider some limit as $s$ approaches

0. The image has 1, 2, and 3 as possible answers.

If $d(s) = \frac{\sin(s)}{s}$ then $d(0)$ is $1$.

Question 7: "$d(s) = \frac{\cot s - 1}{\pi - 2s}$ is continuous at $s = \frac{\pi}{4}$. If $d(\frac{\pi}{4}) = ...$"

Because $d(s)$ is continuous, we have that the function value at the point is the limit of the function as we approach the point. In other words, $d(\frac{\pi}{4}) = \lim_{s \to \frac{\pi}{4}} d(s) = \lim_{s \to \frac{\pi}{4}} \frac{\cot s - 1}{\pi - 2s}$. Let $x = s - \frac{\pi}{4}$. Then $s = x + \frac{\pi}{4}$. As $s \to \frac{\pi}{4}$, we have $x \to 0$. Also $\pi - 2s = \pi - 2(x + \frac{\pi}{4}) = \pi - 2x - \frac{\pi}{2} = \frac{\pi}{2} - 2x$. Furthermore, $\cot(x + \frac{\pi}{4}) = \frac{\cot x \cot(\frac{\pi}{4}) - 1}{\cot x + \cot(\frac{\pi}{4})} = \frac{\cot x - 1}{\cot x + 1} = \frac{\frac{\cos x}{\sin x} - 1}{\frac{\cos x}{\sin x} + 1} = \frac{\cos x - \sin x}{\cos x + \sin x}$.

So $d(\frac{\pi}{4}) = \lim_{x \to 0} \frac{\frac{\cos x - \sin x}{\cos x + \sin x} - 1}{\frac{\pi}{2} - 2x} = \lim_{x \to 0} \frac{\frac{\cos x - \sin x - (\cos x + \sin x)}{\cos x + \sin x}}{\frac{\pi - 4x}{2}} = \lim_{x \to 0} \frac{-2 \sin x}{\cos x + \sin x} \cdot \frac{2}{\pi - 4x} = \lim_{x \to 0} \frac{-4}{\pi - 4x} \cdot \frac{\sin x}{\cos x + \sin x} = \frac{-4}{\pi} \cdot \lim_{x \to 0} \frac{\sin x}{\cos x + \sin x}$. Using L'Hopital's rule, we get $\frac{-4}{\pi} \cdot \lim_{x \to 0} \frac{\cos x}{-\sin x + \cos x} = \frac{-4}{\pi} \cdot \frac{1}{1} = \frac{-4}{\pi}$. However, $\frac{-1}{\pi}$ is one of the choices.

Question 8: "The function $d(s) = \frac{...}{1 - \cos s}$ is discontinuous at $s = ...$"

The expression $\frac{1}{1 - \cos s}$ is undefined when $1 - \cos s = 0$, so $\cos s = 1$. This happens when $s = 2n\pi$ where $n$ is an integer. One possible value is $s=0$.

Question 9: "If the function $d(s) = ... \frac{\sin s}{s + p \sin s}$ is continuous at $s = ...$"

I cannot read the question clearly, but I will assume that $d(s) = \frac{\sin s}{s + p \sin s}$ and the question is to find the value of $p$ such that $d(s)$ is continuous at $s=0$. The $\lim_{s \to 0} d(s) = \lim_{s \to 0} \frac{\sin s}{s + p \sin s}$. Divide numerator and denominator by $s$.

$\lim_{s \to 0} d(s) = \lim_{s \to 0} \frac{\frac{\sin s}{s}}{1 + p \frac{\sin s}{s}} = \frac{1}{1 + p}$. If we want this limit to exist, we require $p \ne -1$. If $p = -1$, we would have $d(s) = \frac{\sin s}{s - \sin s}$ which tends to $1/0$.

Question 10: "$d(s) = ... s + \tan s$. $s> ...$. $... s + 6$ $s \le ...$ "

I assume the question asks for a value of $p$ that makes this function continuous. I assume that we have $d(s) = s + \tan s$ for $s > a$ and $d(s) = ps + 6$ for $s \le a$. To be continuous at $s = a$, we need $a + \tan a = pa + 6$.

Question 11: "The value of $d(1)$ that makes the function $d(s) = ... s^2(1 - \sin s)$ continuous at $s = ...$ is ... "

Question 12: "If $d(s)$ is continuous at $s = p$, $d(p) = 2$, then $\lim_{s \to p} d(s) = ...$"

Since $d(s)$ is continuous at $s=p$, the limit as $s$ approaches $p$ is just the value of the function at $p$. Therefore, $\lim_{s \to p} d(s) = d(p) = 2$.

Question 13: "Which of the following functions is continuous at $s = ...$?"

3. Final Answer

Due to the image quality, some parts are illegible, so providing definite answers is impossible. Here are some educated guesses based on what is visible:

Question 6: Need more information. Assuming $d(s) = \tan(s)$, $p = 0$.

Question 7: $-1/\pi$

Question 8: 0

Question 9: Need more information.

Question 10: Need more information.

Question 11: Need more information.

Question 12: 2

Question 13: Need more information.