statement (2) first, since it's easier:

the y-intercept is fixed, but we can rotate the line so that it has whatever slope we want.

whether the line will hit quadrant II depends on its slope. if the slope is positive or zero, then the line won't hit quadrant II; if the slope is negative, then it will. insufficient.

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statement (1): this is sufficient. in fact, this will be sufficient for ANY value of the slope: it's "yes" for ALL negative slopes, and "no" for slope 0 and for ALL positive slopes.

explanation:

any line with a negative slope goes up to the left, and down to the right, at a constant angle.
forever.

imagine that you have a line with a negative slope, then (ANY negative slope; the actual number -1/6 is immaterial).
pick ANY starting point on this line.
if you go far enough to the left, starting at this point, the line WILL rise into the second quadrant. if the slope of the line is very gentle (i.e., almost flat) and the starting point is way, way deep below the x-axis, then the line might take a REALLY long time to get up above the x-axis, but it will get there.
for the same reason, but going to the right instead, the line WILL also eventually get into the fourth quadrant.

if your line has a positive slope, then, in the same way, it is guaranteed to intersect the first and third quadrants.

make a bunch of sketches if you don't see why this stuff has to be true.

remember, though, that LINES DON'T STOP. for the lines in your sketch with the gentlest and steepest slopes, you may well have to picture imaginary extensions of those lines, waaaaaaaayyyy off the sheet of paper on which you're actually drawing, in order to see how they're going to make it into quadrant 2 (or 4).