Answer
a) Diverging, virtual, $-10.5\;\rm cm$
b) Converging, virtual, $238.4\;\rm cm$
Work Step by Step
a) Since the image is between the lens and the object, it must be a $\bf \color{red}{\bf diverging\; lens}$.
And since the image is in front of the lens, it is a $\color{red}{\bf virtual\; image}$.
Its focal length is given by
$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}$$
$$f=\left[\dfrac{1}{d_o}+\dfrac{1}{d_i}\right]^{-1}$$
Plugging the known;
$$f=\left[\dfrac{1}{37.5}+\dfrac{1}{-8.20}\right]^{-1}=\color{red}{\bf -10.5}\;\rm cm$$
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b) Now the object is between the image and the lens and both of them, the image and the object, are on the same side, so it must be a $\color{red}{\bf converging\; lens}$.
And since the image is in front of the lens, it is a $\color{red}{\bf virtual\; image}$.
Its focal length is given by
$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}$$
$$f=\left[\dfrac{1}{d_o}+\dfrac{1}{d_i}\right]^{-1}$$
Plugging the known;
$$f=\left[\dfrac{1}{37.5}+\dfrac{1}{-44.5}\right]^{-1}=\color{red}{\bf238.4}\;\rm cm$$