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For x and y in $R^2$ we have: $$(x,y)= x_1y_1 + x_2y_2 + kx_1y_2 + kx_2y_1$$ where k is a real parameter. Give an example of orthonormal basis for this scalar product.
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I don't understand this question. Aren't orthonormal basis created in a way to give out the canonical product i.e., the dot product by definition? Can some please elaborate on this?

What you have to do is to come with an orthonormal basis relative to the given inner product, not the usual inner product. You can start with $(1,0)$ and calculate its norm. You will get $|(1,0)|=1$. Then try to find $(x,y)$ such that its inner product with $(1,0)$ is $0$. One such vector is $(x,y)=(1,-frac 1 k)$. Then you have to normalize it so that it becomes a unit vector. You will end up with the orthonormal basis $(1,0), frac 1 c(1,-frac 1 k)$ where $c=sqrt frac 1 k^2-1$. Note the given expression defines an inner product only when $|k|<1$.

You have to find $x,y in mathbb R^2$ such that

1. 
   $(x,x)=(y,y)=1$.

and

2. $(x,y)=0.$

An orthonormal basis is defined with respect to particular inner product. That is, given a vector space $V$ over $mathbb R$, an inner product structure is a map $V times V rightarrow mathbb R$ satisfying certain conditions (bilinearity, symmetry, etc).

In this case, we have $V = mathbb R^2$, and we are given the inner product $$f: V times V rightarrow mathbb R quad f(vec x, vec y) = x_1y_1 + x_2y_2 + k x_1y_2 + kx_2y_1$$

Note that from the inner product, we can define the length/metric as $||vec x|| equiv sqrtf(vec x, vec x)$.

That is, the norm of the vector is the square root of the dot product of a vector with itself.

Now, we are to find two vectors $vec x, vec y in mathbbR^2$ such that:

1. $||x|| = 1$


2. $ ||y|| = 1$


3. $ f(x, y) = 0$

Let's go through the conditions and see what we need.

beginalign*
&||x|| = 1 \
&sqrtf(x, x) = 1 \
&f(x, x) = 1 quad textsquare on both sides\
&x_1x_1 + x_2x_2 + kx_1x_2 + kx_2x_1 = 1 quad textuse $f$ definition\
&x_1^2 + x_2^2 + 2kx_1x_2 = 1 \
endalign*

Similarly, for $||y|| = 1$, we get the condition:
beginalign*
&y_1^2 + y_2^2 + 2ky_1y_2 = 1
endalign*

From the last condition, we get
beginalign*
x_1y_1 + x_2y_2 + k(x_1y_2 + x_2y_1) = 0
endalign*