= Solution
Use the real <Fourier series> convention $g=a_0/2+\sum_{n\geq1}[a_n\cos(n\pi x)+b_n\sin(n\pi x)]$ for period two. Since $g$ is an <even function>, $b_n=0$, and $a_0=\int_{-1}^1x^2dx=2/3$. Twice applying <integration by parts>, with $k=n\pi$, gives
$$
a_n=2\int_0^1x^2\cos(kx)dx
=2\left[\frac{x^2\sin(kx)}k+\frac{2x\cos(kx)}{k^2}-\frac{2\sin(kx)}{k^3}\right]_0^1
=\frac{4(-1)^n}{\pi^2n^2}.
$$
Hence
$$
\boxed{g(x)=\frac13+\frac4{\pi^2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\cos(n\pi x).}
$$
The periodic extension is continuous at its endpoints, so the series takes the specified endpoint values as well as the interior values.
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