Answer
Upright, Virtual, same size, 5.6 cm to the right from the first lens.
Work Step by Step
To find the final image position, we need to find the position of the first image created by the converging lens by using the thin lens law.
$$\dfrac{1}{f_1}=\dfrac{1}{s_1}+\dfrac{1}{s_1'}$$
Thus,
$$s_1'=\left[\dfrac{1}{f_1}-\dfrac{1}{s_1}\right]^{-1}$$
Plugging the known;
$$s_1'=\left[\dfrac{1}{5}-\dfrac{1}{4}\right]^{-1}=\bf -20\;\rm cm$$
This means that the image is on the same side as the object and its distance to the diverging lens is then $s_2=20+12=\bf 32$ cm.
The height of the first image is given by
$$m=\dfrac{h_1'}{h_1}=\dfrac{-s_1'}{s_1}$$
Hence,
$$h_1'=\dfrac{-h_1s_1'}{s_1}$$
Plugging the known;
$$h_1'=\dfrac{-(1)(-20)}{(4)}=\bf5\;\rm cm$$
This first image is the object of the second lens.
This image is at a distance of 32 cm from the second lens, so we need to use the lens thin law again to find the final position of the image.
$$s_2'=\left[\dfrac{1}{f_2}-\dfrac{1}{s_2}\right]^{-1}$$
Plugging the known;
$$s_2'=\left[\dfrac{1}{-8}-\dfrac{1}{32}\right]^{-1}=\bf -6.4\;\rm cm$$
The height of the second image is given by
$$h_2'=\dfrac{-h_1's_2'}{s_2}=\dfrac{-(5)(-6.4)}{(32)}=\bf 1\;\rm cm$$
It is obvious now that the final image is a virtual upright image with the same size as the original object and it is 6.4 cm to the left from the second lens (the diverging one) which is 5.6 cm to the right from the first lens (the converging one).
