$\sqrt{2}$

b) $\sqrt{3}$

c) $\sqrt{5}$

d) $\sqrt{6}$

We can begin by finding the coordinates of the circumcentre and orthocentre of the triangle.

The circumcentre is the intersection point of the perpendicular bisectors of the sides of the triangle. The line passing through the midpoints of AB and AC has a slope of -1 (since the slope of AB is infinity and the slope of AC is 0), and passes through the point ((0+0)/2, (0+2)/2) = (0,1). Using the point-slope form of the equation of a line, this line can be written as y-1 = -1(x-0), or y = -x+1.

Similarly, the line passing through the midpoints of AB and BC has a slope of 1/2 (since the slope of AB is infinity and the slope of BC is -2), and passes through the point ((0+2)/2, (2+0)/2) = (1,1). This line can be written as y-1 = 1/2(x-1), or y = 1/2x+1/2.

The intersection point of these two lines is the circumcentre. Solving for x and y, we get x=2/3 and y=4/3. Therefore, the circumcentre is O (2/3, 4/3).

The orthocentre is the intersection point of the altitudes of the triangle. The altitude from A intersects BC at the point (0,0), so the equation of this altitude is simply x=0. The altitude from B intersects AC at the point ((0+2)/2, (2+0)/2) = (1,1), so the slope of this altitude is -2. Using the point-slope form of the equation of a line, this altitude can be written as y-1 = -2(x-1), or y = -2x+3.

Similarly, the altitude from C intersects AB at the point ((0+2)/2, (0+2)/2) = (1,1), so the slope of this altitude is 1/2. This altitude can be written as y-0 = 1/2(x-2), or y = 1/2x.

To find the intersection point of the altitudes, we need to solve the system of equations:

x = 0

y = -2x+3

y = 1/2x

Substituting x=0 into the second equation, we get y=3. Substituting y=1/2x into the third equation, we get x=2. Therefore, the orthocentre is H (2,3).

The distance between the circumcentre and orthocentre is given by OH = sqrt((2/3 - 2)^2 + (4/3 - 3)^2) = sqrt(2).

Therefore, the answer is (a) sqrt(2).