A hint for part 1

For part 1; Maclaurin expansion $\frac{1}{4-x}= \sum_{k=0}^{\infty }4^{-1-k}x^k$ motivates us to write $ \left| \frac{1}{4}+\frac{x}{16} +\frac{x^2}{64}+\frac{1}{x-4}\right|= \left| \frac{x^3}{64(x-4)} \right|\le \frac{1}{64\cdot 3}<0.01$ for $\left| x \right|\le 1$. Therefore we choose $P(x)= \frac{1}{4}+\frac{x}{16}+\frac{x^2}{64}$ For part 2; The Maclaurin trick fails this time. Therefore we look for real numbers a and b, such that $ \left| f(x) \right|\le 0.01 \forall \left| x \right|\le 1$, where $f(x)= ax+b+\frac{1}{x-4}$. The first derivative of f is $f'(x)= a-\frac{1}{(x-4)^2}$. We observe that $f'(x)=0\Leftrightarrow x= 4- \frac{1}{\sqrt{a}}= x_0$ and this critical point is in the interval $(-1, 1)$ if we choose $a\in (\frac{1}{25}, \frac{1}{9})$. Thinking wisefully about the minimum of the function f, we choose $a= \frac{1}{15}$ so that $f(1)= f(-1) = b- \frac{4}{15}$ The function f is monotonically increasing in $[-1, x_0]$ and monotonically decreasing in $[x_0, 1]$, so $minf= f(1)= f(-1)$ and $maxf= f(x_0)= b+ \frac{4-2\sqrt{15}}{15}$ We need $-0.01\le b-\frac{4}{15}\le \frac{4-2\sqrt{15}}{15}+b\le 0.01$. We choose $a= \frac{1}{15}$ and any value $b\in [\frac{4}{15}- \frac{1}{100}, \frac{1}{100}-\frac{4-2\sqrt{15}}{15}]$ and done.