Using $W = $ {$(x, y) | x, y ∈ R)$} Let $u = (1,3)$ and $v = (2, 1)$ defined by the inner product
$<u, v> = 2v_1u_1 - v_1u_2 - v_2u_1 + v_2u_2$
Find an orthonormal basis and express the vector $w = (1,1)$ as a linear combination with the obtained orthonormal basis.
As the vectors are orthogonal by the definition and equaling $0$, I normalised the vectors with the definition to obtain:
$u = \left(\frac{1}{\sqrt 65}, \frac{3}{\sqrt 65}\right)$ and $v = \left(\frac{1}{2}, \frac{1}{4}\right)$
Assuming this is correct, as it may not be, then normally a linear combination would be formed by
$(1, 1) = c_1\left(\frac{1}{\sqrt 65}, \frac{3}{\sqrt 65}\right) + c_2\left(\frac{1}{2}, \frac{1}{4}\right)$
How do you apply the definition to the linear combination? Anything I try seems awfully messy. Again, perhaps my calculations were wrong but I'm not sure how.
I can't see where you did get $\;\sqrt{65}\;$ for $\;u\;$ :
$$\begin{cases}\left\|u\right\|=\sqrt{2\cdot1\cdot1-1\cdot3-3\cdot1+3\cdot3}=\sqrt5\\{}\\ \left\|v\right\|=\sqrt{2\cdot2\cdot2-2\cdot1-1\cdot2+1\cdot1}=\sqrt5\end{cases}\,\implies$$
$$ w_1:=\frac u{\left\|=\right\|}=\frac1{\sqrt5}(1,3)\,,\;\;w_2:=\frac v{\left\|v\right\|}=\frac1{\sqrt5}(2,1)$$
And now: since we have already an orthonormal basis, expressing any vector as a linear combination of it is pretty simple:
$$(1,1)=\left\langle (1,1),\,w_1\right\rangle\,w_1 +\langle (1,1),\,w_2\rangle w_2$$
where we used $\;\langle,\rangle\;$ to denote the inner product defined by your function