Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)
1a) We can use the mirror equation to find the distance of the image from the mirror:

where

is the focal length

is the distance of the object from the mirror

is the distance of the image from the mirror.
Rearranging the equation, we find

so, the distance of the image from the mirror is

.
1b) The image height is given by the magnification equation:

where

is the heigth of the image and

is the height of the object. By rearranging the equation and using p and q, we find

and the negative sign means the image is inverted.
2) As before, we can find the distance of the image from the mirror by using the mirror equation:

Rearranging it, we find

so, the distance of the image from the mirror is

3) As before, we find the distance of the image from the mirror by using the mirror equation:

Rearranging it, we find

so, the distance of the image from the mirror is

And now we can use the magnification equation to find the image height:

Rearranging it, we find

and the negative sign means the image is inverted.