Expansion of (1-2x)^8:
The expansion of (1-2x)^8 can be found using the binomial theorem. The general term of the expansion is given by:
T(r+1) = C(8,r) * (1)^8-r * (-2x)^r

where C(8,r) is the binomial coefficient, which can be calculated using the formula:

C(n,r) = n! / (r! * (n-r)!)

Numerically Greatest Term:
To find the numerically greatest term, we need to find the value of r that gives the largest value for the term T(r+1). The value of r that gives the largest value for T(r+1) is the one that satisfies the inequality:

|T(r+1)| > |T(r)| and |T(r+1)| > |T(r+2)|

We can simplify this inequality by dividing both sides by |T(r+1)|:

1 > |T(r)| / |T(r+1)| and |T(r+2)| / |T(r+1)| < />

We can rewrite the first inequality as:

|T(r)| / |T(r+1)| < />

Taking the absolute value of both sides, we get:

|T(r)| < />

This means that the term T(r+1) is numerically greater than T(r).

Substituting x = 2:
Substituting x = 2 in the general term, we get:

T(r+1) = C(8,r) * (1)^8-r * (-2*2)^r
T(r+1) = (-1)^r * C(8,r) * 2^r

Finding the Numerically Greatest Term:
To find the numerically greatest term, we need to evaluate the absolute value of the general term for different values of r, and find the value of r that gives the largest absolute value.

The values of T(r+1) for r = 0 to 8 are:

T(1) = 12870
T(2) = -30080
T(3) = 35960
T(4) = -27392
T(5) = 13888
T(6) = -4576
T(7) = 1008
T(8) = -128
T(9) = 1

The term with the largest absolute value is T(3), which is 35960. Therefore, the numerically greatest term in the expansion of (1-2x)^8, when x = 2, is:

T(3) = (-1)^2 * C(8,2) * 2^2
T(3) = 28 * 4 * 35960
T(3) = 4020736