(i) True.
Let a+b=9n where n is an integer and a and b are integers. So b=9n-a.
10a+b=10a+9n-a=9(a+n). Therefore a+n=(10a+b)/9, and, since a, n and b are integers, 10a+b must be divisible by 9 when the sum of the digits is a multiple of 9. The argument can be extended to cases where a and b are positive and less than 10. In cases where a>10, the argument can be extended to consider a=10a₁+a₂ where a₁ and a₂ are less than 10. So in any integer of n digits, their sum is divisible by 9 if the entire integer is divisible by 9.
(ii)(a) True. (The sum of the digits is divisible by 3)
Let a+b=3n, then 10a+b=10a+3n-a=3(3a+n). 3a+n=(10a+b)/3 so 10a+b is exactly divisible by 3 when a, b and n are integers. See (i) for extension of the argument.
(ii)(b) False, ABC must have C=0 or 5, because even multiples of 5 can be represented by 10x and odd multiples by 10x+5 where x is an integer (2x is an even multiple, 2x+1 is an odd multiple).
(iii) False. A n-sided polygon has n internal angles.
Two non-parallel lines form one internal angle between them. Three open lines form an additional internal angle. (As many external angles are also formed such that the sum of any internal and associated external angle is 360 degrees.) In general, n open lines create n-1 internal angles, because one angle is created between each pair of adjacent lines. When the end lines are connected by another line, two angles are formed, one at each end of the line. The figure created is a polygon, n-gon. If the original number of lines was n, the number of sides of the polygon increases by 1 to n+1 and the number of angles increases by 2 to n-1+2=n+1, the number of sides.