The parts of the given question are solved below:

i) The equation we have is: 5p + 2 = 17

Note: To find solution of the hit and trial method, we check for common values like -1, -2, -3 , 0 , 1 2, 3 and so on.

Lets, check for p = 1

Then, L.H.S = 5(1) + 2 = 7 ≠ RHS

So, p = 1 is not a solution of the given equation.

Now,

Check for p = 2

L.H.S = 5(2) + 2 = 12 ≠ RHS

So, p = 2 is not a solution of the given equation.

Now,

Check for p =3

L.H.S = 5(3) + 2 = 17 = RHS

So, p = 3 is a solution of the given equation.

ii) The equation we have is 3m – 14 = 4

Note: To find solution of the hit and trial method, we check for common values like -1, -2, -3, 0 , 1 2, 3 and so on.

Let us check for m = 2

Then, LHS = 3(2) – 14 = 6 – 14 = -8 ≠ RHS

So, m = 2 is not a solution of the given equation.

Now,

Let us check for m = 3

Then, LHS = 3(3) – 14 = 9 – 14 = -5 ≠ RHS

So, m = 3 is not a solution of the given equation.

Now,

Check for m = 5

LHS = 3(5) – 14 = 15 – 14 = 1 ≠ RHS

So, m = 5 is not a solution of the given equation.

Now,

Check for m = 6

LHS = 3(6) – 14 = 18 – 14 = 4 = RHS

Hence,

LHS = RHS

Therefore, m = 6 is the solution of the given equation.