No, f can not be 1-to-1. The reason is because the domain has dimension 3 but the co-domain has dimension of 2. A mapping can not be 1-to-1 when the the dimension of the domain is greater than the dimension of the co-domain.

No, f cannot be onto. The reason is because the dimension of the domain (2) is less than the dimension of the codomain(3).

For a function to be onto, the dimension of the domain must be less than or equal to the dimension of the codomain.

f cannot be onto. The reason is because the domain, V, has a dimension less than the dimension of the codomain, W.

f can be 1-to-1 since the dimension of V is less-than-or-equal to the dimension of W. However, just because f can be 1-to-1 based off its dimension does not mean it is guaranteed. 

f preserves both vector addition and scalar multiplication because it was stated to be a homomorphism in the problem statemenet. The definition of a homomorphism is a mapping that preserves both vector addition and scalar multiplication.

f is a homomorphism because it preserves both vector addition and scalar multiplication.

To show this we need to prove both statements

Proof f preserves vector addition

Let u and v be arbitrary vectors in with the form and

Consider . Applying the definition of f we get

This is the same thing as

 Hence, f preserves vector addition because

Proof f preserves scalar multiplication

 Let u be an arbitrary vector in with the form and let k be an arbitrary real constant. 

Consider

This is the same thing we get if we consider

Hence f preserves scalar multiplication because for all vectors u and scalars k.

f is not onto. This is because not every vector in is in the image of f. For example, the vector is not in the image of f. Hence, f is not onto.

We could also see this quicker by looking at the dimension of the domain and codomain. The domain has dimension 2 and the codomain has dimension 3. A mapping can't be onto and have a domain with a lower dimension than the codomain.

Finally, we know f preserves vector addition and scalar multiplication because it is a homomorphism.

To find the null space consider the equation

This gives a system of equations

The only solution to this system is 

Thus the null space consists of the single vector

To find the rank of , we first find the image of . The image of  is any vector of form  where a is any real number. This is a line in . Therefore the image of  is a 1 dimensional subspace. Thus the answer is 1.

The image is the space spanned by the vectors  and . The image has a basis of  vectors. Therefore the image has dimension .  Thus the rank is .

The maximum possible rank of a function is the dimension of the domain. The domain of  is . Therefore the maximum possible rank of  is .

The null space is the subspace such that  maps to the zero vector in the codomain. The largest possible null space is when the entire domain goes to the zero vector. The domain of  is  . Therefore the largest possible null space would be   which has a dimension of . Thus the largest possible dimension for the null space is .