I'm assuming you're asked to find the equations for lines that fit the given descriptions.
In the first case, having no x-intercept means that the line must be horizontal and not [tex]x=0[/tex], so it will be entirely determined by the y-intercept, which is -5. So the line has equation [tex]y=-5[/tex].
In the second case, the line is parallel to [tex]4y-2x=7[/tex], or equivalently [tex]y=\dfrac12x+\dfrac74[/tex], which has slope [tex]\dfrac12[/tex]. Parallel lines have the same slope. With the point-slope formula, you can find its equation:
[tex]y-(-7)=\dfrac12(x-2)\implies y=\dfrac12x-8[/tex]
In the third case, the line is perpendicular to [tex]-7+3y=6x[/tex], or [tex]y=2x+\dfrac73[/tex]. Perpendicular lines have slopes that are negative reciprocals of one another. Here, the given line has slope 2, which means the slope of its perpendicular counterpart is [tex]-\dfrac12[/tex]. Point-slope formula again:
[tex]y-6=-\dfrac12(x-(-3))\implies y=-\dfrac12x+\dfrac92[/tex]
In the fourth case, you first need to find the slope of such a line. Note that the x-coordinates are the same, which means the line will be vertical and the equation is determined entirely by the x-intercept. So the equation is [tex]x=7[/tex].
