Solution:

(i) Let the co-ordinates of P(x, y) divides AB in the ratio m:n.

A(3,2) and B(5,1) are the given points.

Given m:n = 1:2

x₁ = 3 , y₁ = 2 , x₂ = 5 , y₂ = 1 , m = 1 and n = 2

By Section formula x = (mx₂+nx₁)/(m+n)

x = (1×5+2×3)/(1+2)

x = (5+6)/3

x = 11/3

By Section formula y = (my₂+ny₁)/(m+n)

y = (1×1+2×2)/(1+2)

y = (1+4)/3

y = 5/3

Given P lies on the line 3x-18y+k = 0

Substitute x and y in above equation

3×(11/3)-18×(5/3)+k = 0

11-30+k = 0

-19+k = 0

k = 19

Hence the value of k is 19.

(ii) Let the co-ordinates of P(x, y) divides AB in the ratio m:n.

A(3,-5) and B(-4,8) are the given points.

Given AP/PB = k/1

m:n = k:1

x₁ = 3 , y₁ = -5 , x₂ = -4 , y₂ = 8 , m = k and n = 1

By Section formula x = (mx₂+nx₁)/(m+n)

x = (k×-4+1×3)/(k+1)

x = (-4k+3)/(k+1)

x = (-4k+3)/(k+1)

By Section formula y = (my₂+ny₁)/(m+n)

y = (k×8+1×-5)/(k+1)

y = (-4k+3)/(k+1)

Co-ordinate of P is ((-4k+3)/(k+1), (8k-5)/(k+1))

Given P lies on line x+y = 0

Substitute value of x and y in above equation

(-4k+3)/(k+1) + (8k-5)/(k+1) = 0

(-4k+3) + (8k-5) = 0

4k-2 = 0

4k = 2

k = 2/4 = ½

Hence the value of k is ½ .