whew! this one is a serious cacophony of rephrasing and interpretation.

let's translate:

m is not a factor of p.

if m were a factor of p, then the remainder upon dividing p by m would be 0.

therefore, we can translate the above statement as follows:
"the remainder upon dividing p by m is not 0."
in other words, it's an integer greater than 0.

--

the question:

is r > 1 ?

here's a HIGH-LEVEL INTERPRETATION of this problem.

if the remainder WERE 1, then p would be 1 more than a multiple of m.
if this is the case, then p and m CANNOT have any common factors, other than 1. (this is so because all factors of m are factors of (p-1), which is a multiple of m; a number greater than 1 can't be a factor of both (p-1) and p, which are consecutive integers.)

therefore, if m and p have common factors, then the answer to this question is YES.

(note that the converse is not necessarily true: even if there are no common factors, the answer still could be yes. for instance, 17 divided by 6 leaves a remainder of 5, even though 17 and 6 have no common factors. but, if we can establish that there are common factors, then that's enough to show that the answer is Yes.)

--

statement (1)

if this is true, then m and p have the factor 2 in common, so, YES.
sufficient.

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statement (2)

this doesn't tell you whether m and p have common factors.
if m = 5 and p = 6, for instance, then r = 1.
if m = 10 and p = 15, then r = 5, which is > 1.
insufficient.

ans (a)

--

secondary solution: (because the primary solution above is fairly obnoxious)

JUST PLUG IN NUMBERS.


statement (1)

let's just PICK A WHOLE BUNCH OF NUMBERS WHOSE GCF IS 2 and watch what happens. let's try to make the numbers diverse.
say,
4 and 6
6 and 8
8 and 10
10 and 12
...
4 and 10
6 and 14
6 and 16
8 and 18
8 and 22
...
in all nine of these examples, the remainders are greater than 1. in fact, there is an obvious pattern, which is that they're all even, since the numbers in question must be even.

in fact, i just thought of this, which is a much nicer, more ground-level approach to statement one:
in statement 1, both m and p are even. therefore, the remainder is even, so it's greater than 1.

done.

sufficient.

--

statement (2)
just pick various numbers whose lcm is 30.
notice the numbers selected above:
5 and 6 --> remainder = 1
10 and 15 --> remainder = 5 > 1
insufficient.