Regarding your first question,
All evolved stars do emit like black bodies, but this is not true for young stellar objects which may exhibit a large infrared excess.
Thus if you consider, for example, a main sequence star (which are the most common stars) and you measure its mass $M$ and radius $R$, you can indeed infer its spectral properties. Firstly, let us write the equation for the luminosity $L$ of the star:
$L = 4 \pi R ^ 2 F$
Where $F$ is the spectral flux density at the surface of the object.
If you assume that the emission can be described by the black body model, then you have:
$L = 4 \pi R ^ 2 \sigma T ^ 4$
Where $\sigma$ is the Stefan-Boltzmann constant and $T$ the temperature.
Now you need a mass-luminosity function to relate the luminosity $L$ to the mass $M$ that you have measured. This relation can be derived theoretically, by thermodynamic considerations, and yields this kind of equation:
$L \propto M ^ a$
In the case of main sequence stars, this relation is well known and the value $a = 3,5$ is commonly used.
Thus, using the previous equation, it is possible to determine the surface temperature of our main sequence star:
$T = \left( \dfrac{M ^ {3,5}}{4 \pi R ^ 2 \sigma }\right) ^ {1/4}$
Then, as you said, you can benefit from Wien's displacement law to find the wavelength $\lambda_{max}$ of the emission peak in the black body spectrum, given the temperature $T$ that you just found, using:
$\lambda_{max} = \dfrac{b}{T}$
By doing so you have determined the spectral type of the star only from $M$ and $R$. Be warned however that this depends on the choice of the mass-luminosity relation.
Note: Additionnaly, having determined the luminosity $L$ and temperature $T$ of the star would allow you to place it on a Hertzsprung–Russell diagram.