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In exercise 1 points 1) and 2) are proved by induction on $n$. Point 3 follows from 2) by solving the inequality $$\frac{1}{2^{1+n/2}}<\frac{1}{200}$$.
In exercise 2, point 1) the answer is yes because symmetric difference is associative, commutative and has the property $$A\triangle A=\emptyset$$. So the effect of taking a symmetric difference with $A$ can be canceled.
In exercise 3 you can draw the graph of $f$ on
Desmos. In fact, it is easy to draw the graph by hand because $$f(x)=\begin{cases}x^2+x,&x\ge0\\x^2-x,&x<0\end{cases}$$. It is also clear that $f(x)$ is even, i.e., $f(-x)=f(x)$, so it is sufficient to study $f(x)$ for $x\ge0$.
If $g(x)$ is the restriction of $f(x)$ to $\mathbb{R}^+$, then it is clear from the graph that $g$ is a bijection from $\mathbb{R}^+$ to $\mathbb{R}^+$. The inverse $g^{-1}(y)$ is found by solving the equation $x^2+x=y$ for $x$.
In exercise 4 the elements of $A$ can be easily enumerated: $A=\{1,2,3,4,5\}$ (if $\mathbb{N}$ starts from 1). To find if any of them also belong to $B$ we can compute $$\frac{n^2-16}{n-2}$$ for these $n# to see if the result is an integer.
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