 Solution

For any $q\in {\Pi }_n$, write $f-q=(f-p)+(p-q)$. The stated orthogonality makes the two summands perpendicular in the given inner product, because $p-q\in {\Pi }_n$. The Pythagorean theorem for inner product spaces therefore givesEquality is attained at $q=p$, and positive definiteness makes it unique. Thus $p$ is the orthogonal projection of $f$ onto the polynomial subspace.

For the weighted example set $p(x)=A+Bx$. Orthogonality against the basis $1,x$ givesOdd terms vanish on the symmetric interval, so $A=3/5$ and $B=0$. ConsequentlyThe weight $x^2$ vanishes only at one point; its integral against $g^2$ is positive for every nonzero continuous $g$, so it really does define an inner product. For an optional error check, the squared minimum error is $2/7-12/25+6/25=8/175$.