Answer: \(\left(\frac12,\frac14\right)\) and \((2,1)\).

From

\(x^2=5y-1,\)

we get

\(y=\frac{x^2+1}{5}.\)

At an intersection point,

\(\frac{x^2+1}{5}=x^2-2x+1.\)

Multiplying by 5 gives

\(x^2+1=5x^2-10x+5.\)

So

\(4x^2-10x+4=0,\)

or

\(2x^2-5x+2=0.\)

Factorise:

\((2x-1)(x-2)=0.\)

Thus

\(x=\frac12\quad\text{or}\quad x=2.\)

When \(x=\frac12\),

\(y=\frac{(\frac12)^2+1}{5}=\frac14.\)

When \(x=2\),

\(y=\frac{2^2+1}{5}=1.\)

Therefore the points of intersection are

\(\boxed{\left(\frac12,\frac14\right)\text{ and }(2,1)}.\)