I want to verify if I am thinking of parts b) and c) correctly.
To find its kernel, we must find the nullspace of the matrix equation
$ \begin{bmatrix} 5^5 & 5^4 & 5^3 & 5^2 & 5 & 1 \\ 4^5 & 4^4 & 4^3 & 4^2 & 4 & 1 \\ 3^5 & 3^4 & 3^3 & 3^2 & 3 & 1 \\ 2^5 & 2^4 & 2^3 & 2^2 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_5 \\ a_4 \\ a_3 \\ a_2 \\ a_1 \\ a_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} $
Due to the fact that this is a vandermonde matrix with distinct geometric ratios for its rows, this matrix has full rank and thus has a trivial nullspace I think?
Then, to show part c that there is a unique polynomial such that $f(j) = a_j$, can we then just say that $ \begin{bmatrix} 5^5 & 5^4 & 5^3 & 5^2 & 5 & 1 \\ 4^5 & 4^4 & 4^3 & 4^2 & 4 & 1 \\ 3^5 & 3^4 & 3^3 & 3^2 & 3 & 1 \\ 2^5 & 2^4 & 2^3 & 2^2 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a_5 \\ a_4 \\ a_3 \\ a_2 \\ a_1 \\ a_1 \end{bmatrix} = \begin{bmatrix} a_5 \\ a_4 \\ a_3 \\ a_2 \\ a_1 \\ a_0 \\ \end{bmatrix} $
has a unique solution due to nontrivial nullspace?
Well, you are right saying that this is a Vandermonde matrix and hence is invertible, however, this exercise is basically how one proves this fact. You can use the formula to compute such a determinant (which does not use this exercise), but I guess this is not the spirit of the assignment.
How about saying instead that $T$ is invertible, since if $f\in V$ is such that $T(f)=0$, then $f$ has $6>5$ distinct roots, so that $f=0$.
For the rest of the exercise, your solution is fine.