Solving for the determinant only given one column of values.

You're right about the first question. For the second one, note that\begin{align}\det\begin{bmatrix} a & 4 & d \\ b & 5 & e \\ c & 6 & f \\ \end{bmatrix}&=\det\begin{bmatrix} a & 3+1 & d \\ b & 3+2 & e \\ c & 3+3 & f \\ \end{bmatrix}\\&=\det\begin{bmatrix} a & 3 & d \\ b & 3 & e \\ c & 3 & f \\ \end{bmatrix}+\det\begin{bmatrix} a & 1 & d \\ b & 2 & e \\ c & 3 & f \\ \end{bmatrix}.\end{align}

Start with properties of matrices:

$$\text{det}\begin{bmatrix}a_1+d_1 & b_1 & c_1 \\ a_2 + d_2 & b_2 & c_2 \\ a_3+d_3 & b_3 & c_3\end{bmatrix} = \text{det}\begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix} + \text{det}\begin{bmatrix}d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3\end{bmatrix}$$

This is true for any column. It is easy to show that:

$$\text{det}\begin{bmatrix}a & 8 & d \\ b & 8 & e \\ c & 8 & f\end{bmatrix} = 8\cdot \text{det}\begin{bmatrix}a & 1 & d \\ b & 1 & e \\ c & 1 & f\end{bmatrix}$$

$$\text{det}\begin{bmatrix}a & 4 & d \\ b & 5 & e \\ c & 6 & f\end{bmatrix} = \text{det}\begin{bmatrix}a & 1 & d \\ b & 2 & e \\ c & 3 & f\end{bmatrix}+3\cdot \text{det}\begin{bmatrix}a & 1 & d \\ b & 1 & e \\ c & 1 & f\end{bmatrix}$$