Task 3. Find those real values of the parameter $a$ for which the inequality

$$
x^{2}-(a+1) x+a+1>0
$$

is true for all values of $x$ satisfying the condition $|x| \leqslant 1$.
First solution. Solve the inequality (1) with respect to the parameter:

$$
\begin{aligned}
& x^{2}-a x-x+a+1>0 \\
& x^{2}-x+1>a x-a \\
& a(x-1)\frac{x^{2}-x+1}{x-1}=\frac{x(x-1)+1}{x-1}=x+\frac{1}{x-1}
\end{aligned}
$$

We construct the graph of the function $a(x)=x+\frac{1}{x-1}$ on $-1 \leqslant x \leqslant 1$.

For this, we compute:

$$
a^{\prime}(x)=1-\frac{1}{(x-1)^{2}}=0
$$

The root of this equation, belonging to the half-interval $[-1 ; 1]$, is only 0. We compute: $a(0)=-1$; $a(-1)=-1.5 ; a(1)=-\infty$.

Using the graph, we get the answer: $-1<a<+\infty$.

Second solution. The graph of the function $a(x)$ can be constructed using a calculator (and without the help of a derivative!). Using a calculator, it is easy to discover that the maximum value of the function $a(x)$ is reached at the point $x=0$. After this, it remains to prove that

$$
x+\frac{1}{x-1} \leqslant-1, \text { if }-1 \leqslant x \leqslant 1
$$

And this is already a very simple matter.