Let $f(x)=\frac{\sin(x)}{x}$ when $x\neq 0$ and $f(x)=1$ when $x=0$. Starting with the Taylor polynomial of degree $2n+1$ for $\sin(x)$ and the estimate for the remainder term, show that
$f(x)=(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots +(-1)^n \frac{x^{2n}}{(2n+1)!})+R_{2n,0,f}(x)$ where $|R_{2n,0,f}(x)|\leq \frac{|x|^{2n+1}}{(2n+2)!}$ and use this to conclude that $\int_0^1 f \approx \int_0^1 (1-\frac{x^2}{3!}+\frac{x^4}{5!})dx = \frac{1703}{1800}\approx .946$ with an error of less than $10^{-3}$.
The first part of this seems pretty straightforward unless I'm missing something with the piecewise nature of $f$. The second statement is not immediately clear to me, though.
Basically you estimate $ f (x) $ with the first 3 terms of it's expansion. There will be an error involved, whose value is given in your question. Since we are taking the expansion up to the $ n=2$ term, our remainder is
$ R \leq \frac {x^5}{6!}$
To estimate$ \int_0^1 f (x) dx$ we take the integral of the first three terms of the expansion (which is easily computed). The error involved then is
$\displaystyle\int_0^1 \frac {x^5}{6!} = \frac {1}{6(6!)} = \frac {1}{4320} \leq 10^{-3}$ So you have your second part