There are a lot of problems here. I'll do the first four to get you started.
Problem 1
The equation y = 2x-5 is in the form y = mx+b with m = 2 as the slope and b = -5 as the y intercept. As long as we keep the slope the same, but change the y intercept, then we can form any parallel line we want. So for example, one answer (of infinitely many) is y = 2x+7. This line still has a slope of 2, but the y intercept is now 7.
One possible y = 2x+7
Problem 2
The point (a,1) means x = a and y = 1. Plug these items into the equation y = 3x-5 and solve for 'a'
y = 3x-5
1 = 3a-5
1+5 = 3a ... adding 5 to both sides
6 = 3a
3a = 6
a = 6/3 dividing both sides by 3
a = 2 is the answer
Problem 3
Let's solve the second equation for y
x-3y+7 = 0
x-3y = 7 ... adding 7 to both sides
-3y = -x+7 subtract x from both sides
y = (-x+7)/(-3) divide both sides by -3
y = (-x)/(-3) + 7/(-3)
y = (1/3)x - 7/3
We only need to worry about the slope here, which is 1/3
The slope of the first equation given (y = -3x+1) is -3
The fact the two slopes -3 and 1/3 multiply to -1 is enough information to prove the two lines are perpendicular. Put another way, the two slopes are negative reciprocals of one another. The idea is to flip the fraction and the sign from positive to negative. It might help to think of -3 as -3/1.
Problem 4
y = -2
The x axis is horizontal, so the answer is also horizontal as it is parallel to the x axis. We want the answer to go through (3,-2) meaning that we want all points on this horizontal line to have their y coordinate fixed at -2. The x coordinate can be anything you want. Two points on this line are (0,-2) and (3,-2)
Note that x does not show up in the answer at all, which helps reinforce the idea that x can be anything you want and it won't matter as the y value is always -2. You can think of y = -2 as y = 0x+(-2) and see the slope here is 0.
