A very ad-hoc solution to Puzzle 1 in nine lines (technically ten, because you have to extend one past the point you anchor it to):
((0,0) is the origin)
Draw a line from (-2,2) to (-1,-1). Mark (4/3,0). Draw a line from A to that point and mark its intersection with the opposite side of the triangle. Call it B.
Draw a line from (0,0) to (1,-2). Mark (3/2,-3). Draw a line through that point and (1,3), and mark (13/12,2). Call it C.
Draw a line from (0,-2) to (2,-1). Mark (3,-1/2). Draw a line through that point and (1,3), and mark (2,5/4). Draw a line through that point and (-1,1), and mark (0,13/12). Call it D.
BC and BD are the required other sides.
As for Puzzle 2, let's adapt Jens's idea, but abuse some grid lines to cheapen the construction.
Our B line will be the one through (-1,0), which we don't even need to draw: its intersection points are (-1,0) and (0,1).
Our A line is the one through (0,-1). Draw it, and draw the line connecting the upper intersection to (-1,0). It intersects the y-axis at D.
Our C line is the one through (1,0). Draw it, and draw the line connecting the lower intersection to (0,1). It intersects the x-axis at E.
DE is our Pascal line, and two more lines (for a total of 7) give us the tangents!
The Pascal line is in green, and there are no tangents because the lower one would be too close to the other lines (and the thicker lines cover for inaccurate drawing on mobile!)
EDIT: "You can save one more line."
Let's just skip straight to the Pascal line, which is the line from (-1,1) to (2,-1), and save four!