A) By theorem 2.1. $f(x) = x^3$ is uniformly continuous on $[0,1]$. On $(0,\infty)$ on the other hand, by Satz 2.12., it is not uniformly continuous. To see this, pick some $h > 0$. Then $|x^3 + 3x^2 h + 3xh^2 + h^3 -x^3|=|3x^2 h + 3xh^2 + h^3| \ge 3x^2h$ which is unlimited in $x$.
Geometrically this is so because the slope of $x^3$ gets arbitrarily large as $x$ gets large.
B) Let $f(x)= \dfrac{1}{\sin x} - \dfrac{1}{x} = { x - \sin x\over x \sin x}$. Then $f$ is continuous for $x \in (0,1]$ because $x, \sin x$ are continuous and sums, differences and quotients of continuous functions are continuous (if the denominator is non-zero which is the case here).
Note that $\lim_{x \to 0}f(x) = 0$. (to see this apply de l'Hôpital's rule) Hence we may continuously extend $f$ to the closed and bounded interval $[0,1]$ (by defining $f(0):= 0$). It then follows from theorem 2.1. that $f$ is uniformly continuous on $[0,1]$ and since $(0,1) \subseteq [0,1]$ also on $(0,1)$.
C) Let $f(x) = \dfrac{1}{1+x^2}$. Then $f'(x) = -\dfrac{2x}{(1+x^2)^2}$. This function is uniformly continuous on $[-1,1]$ (by theorem 2.1) and therefore bounded on $[-1,1]$. Outside $[-1,1]$ we have
$$\left | -\dfrac{2x}{(1+x^2)^2} \right | = \left | -\dfrac{2x}{(1+x^2)^2} \cdot {{1\over x} \over {1 \over x}}\right | = \left | -\dfrac{2}{((1+x^2){1\over \sqrt{x}})^2} \right | = {2 \over (1+x^{3/2}))^2}\le {1\over 2}$$
Here you want to apply Satz 2.2. and to this end you want to show that $f'$ is bounded. How you do this depends on the function in question but in this case you can achieve this by dividing the domain into two sets and showing that $f'$ is bounded on each and hence also on the union.
Why did I "multiply by ${1 \over x}$"? Well, I wanted to get a bound on the expression and thought about how to eliminate $x$ from the numerator. So I tried multiplying by $1={{1\over x} \over {1 \over x}}$ to see if it did the trick and it did. Once you have a constant in the numerator it's easy to see that it's bounded for $|x|$ greater some non-zero constant.
D) Apply Satz 2.2. to $g(x) = \sin x$ to deduce that it is uniformly continuous.
For the last one you can apply Satz 2.12. This is a bit tricky, you can find a full solution here.
For $r>0$ let $s_r\in (0,p)$ such that $\forall x,y\in [0,2p]\;(|x-y|<s_r\implies |f(x)-f(y)|<r). $
For any $x,y\in \mathbb R$ with $|x-y|<s_r$ there exists $n\in \mathbb Z$ such that $\{x,y\}\subset [np, (n+2)p].$ Because if $n=\max \{m\in \mathbb Z: mp\leq \min (x,y)\}$ then $\min (x,y)<(n+1)p,$ so $$np\leq \min (x,y)\leq\max (x,y)<\min (x,y)+s_r<(n+1)p+s_r<(n+2)p .$$
Now $\{x-np,y-np\}\subset [0,2p]$ and $|(x-np)-(y-np)|=|x-y|<s_r.\;$ Therefore $$|f(x)-f(y)|=|f(x-np)-f(y-np)|<r.$$
Best Answer
Continuous functions map compact sets to compact sets. Thus the image of $[a,b]$ will be some closed interval $[\alpha, \beta]$. As $f(x) > 0$ for all $x$, it must be that $\alpha > 0$.
Stated differently: continuous functions on compact sets attain their infimum and supremum values. Since $f(x) > 0$, the infinimum must be bigger than $0$. Denote it by $\alpha$.