Let's begin by solving the equations of the straight lines in the form $y = mx + c$ (slope-intercept form), based on the problems listed in the image.

First Set of Problems:

1. Find the equation of the straight line that:

   a) Has a gradient (slope) of 3 and passes through (-3, 3):

   We use the point-slope form: $y - y_1 = m(x - x_1)$ where $m = 3$, $(x_1, y_1) = (-3, 3)$.

   Substituting: $y - 3 = 3(x + 3)$ Simplifying: $y = 3x + 9 + 3$ Final equation: $y = 3x + 12$

   b) Has a gradient of -0.5 and passes through (0, 0):

   Using the same formula with $m = -0.5$ and $(x_1, y_1) = (0, 0)$: $y - 0 = -0.5(x - 0)$ Simplifying: $y = -0.5x$

   c) Is parallel to $2y = 5x + 7$ and passes through (0, -3.5):

   First, rearrange $2y = 5x + 7$ into slope-intercept form $y = mx + c$: $y = \frac{5}{2}x + \frac{7}{2}$ The slope is $\frac{5}{2}$, and since the lines are parallel, their slopes are equal. Using point (0, -3.5): $y + 3.5 = \frac{5}{2}(x - 0)$ Simplifying: $y = \frac{5}{2}x - 3.5$

   d) Is parallel to $9x - 5y = -3$ and passes through (5, -3):

   First, rearrange $9x - 5y = -3$ to get it in slope-intercept form: $-5y = -9x - 3$ $y = \frac{9}{5}x + \frac{3}{5}$ The slope is $\frac{9}{5}$, and using the point (5, -3): $y + 3 = \frac{9}{5}(x - 5)$ Simplifying: $y = \frac{9}{5}x - 12$

Second Set of Problems:

2. Find the equation of the straight line joining:

   a) A is (-2, 1) and B is (-1, 4):

   The slope $m$ is given by: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{-1 - (-2)} = \frac{3}{1} = 3$ Using point (-2, 1): $y - 1 = 3(x + 2)$ Simplifying: $y = 3x + 7$

   b) A is (-3, 4) and B is (6, 1):

   The slope $m$ is: $m = \frac{1 - 4}{6 - (-3)} = \frac{-3}{9} = -\frac{1}{3}$ Using point (-3, 4): $y - 4 = -\frac{1}{3}(x + 3)$ Simplifying: $y = -\frac{1}{3}x + 3$

   c) A is (-2, 1) and B is (3, 1):

   Since the y-coordinates are the same, the line is horizontal: $y = 1$

Third Set of Problems:

3. A is (-10, 5) and B is (8, -4):

   a) The slope $m$ is: $m = \frac{-4 - 5}{8 - (-10)} = \frac{-9}{18} = -\frac{1}{2}$ Using point (-10, 5): $y - 5 = -\frac{1}{2}(x + 10)$ Simplifying: $y = -\frac{1}{2}x$

I have solved the initial parts. Would you like further assistance on this?