**Given:** In triangle ABC, right-angled at B, tan A = 1/√3

**To find:**
(i) sin A cos C cos A sin C
(ii) cos A cos C - sin A sin C

**Solution:**
(i) We know that sin A = opposite/hypotenuse = BC/AC and cos C = adjacent/hypotenuse = AB/AC

Thus, sin A cos C = BC/AC * AB/AC = BC * AB / AC^2

Also, cos A = adjacent/hypotenuse = AB/AC and sin C = opposite/hypotenuse = BC/AB

Thus, cos A sin C = AB/AC * BC/AB = BC/AC

Therefore, sin A cos C cos A sin C = (BC * AB / AC^2) * (BC/AC) = BC^2 * AB / AC^3

Now, we need to find the values of BC, AB, and AC.

In a right-angled triangle, if one acute angle is A, then the other acute angle is 90° - A.

Thus, in triangle ABC, angle C = 90° - A.

We know that tan A = opposite/adjacent = BC/AB

Therefore, BC = AB * tan A = AB * (1/√3) = AB/√3

Using Pythagoras theorem, we have AC^2 = AB^2 + BC^2

Substituting the values of BC, we get AC^2 = AB^2 + (AB/√3)^2 = 4AB^2/3

Thus, AC = AB/√3 * 2/√3 = 2AB/3

Substituting the values of BC, AB, and AC, we get:

sin A cos C cos A sin C = (BC^2 * AB) / AC^3 = (AB^2/3 * AB) / (2AB/3)^3 = 3/8

Therefore, sin A cos C cos A sin C = 3/8

(ii) cos A cos C - sin A sin C = AB/AC * AB/AC - BC/AC * BC/AB

Substituting the values of BC, AB, and AC, we get:

cos A cos C - sin A sin C = (AB/AC)^2 - (AB/AC) * (AB/√3) * (√3/AB) = (AB/AC)^2 - 1/3

We know that AC = 2AB/3

Therefore, cos A cos C - sin A sin C = (AB/(2AB/3))^2 - 1/3 = 1/4 - 1/3 = -1/12

Thus, cos A cos C - sin A sin C = -1/12

Hence, the values of sin A cos C cos A sin C and cos A cos C - sin A sin C are 3/8 and -1/12 respectively.