For part (a), this is just development (Laplace expansion) of the determinant by the first row. Actually the $\det()$ factors should have alternating signs. Since the only occurrences of $x$ are in that first row, all the $\det()$ expressions are constants, and one gets a polynomial of degree at most $n-1$ (from the final term) in $x$.
For part (b), that $P(a_i)=0$ for $i=2,3,\ldots,n$ is just the fact that $P(a_i)$ equals the determinant of the matrix obtained by substituting $a_i$ for $x$, so from the original matrix $a_1$ has been replaced by $a_i$, and as this matrix has its rows $1$ and $i$ identical, its determinant vanishes. all this uses is that the Laplace expansion used commutes with such substitution. Furthermore a polynomial of degree at most $n-1$ with $n-1$ specified roots $a_2,\ldots,a_n$ can only be a scalar multiple of $(x-a_2)\ldots(x-a_n)$.
For part (c), this is just remarking that the $\det()$ in question is $(-1)^{n-1}$ times the determinant of the lower-left $(n-1)\times(n-1)$ submatrix, which determinant precisely matches the definition of $V_{n-1}(a_2,\ldots,a_n)$.
For part (d) write $(-1)^{n-1}\prod_{i=2}^n(x-a_i)=\prod_{i=2}^n(a_i-x)$ to get
$$
V(x,a_2,\ldots,a_n)=
(-1)^{n-1}V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(x-a_i)
=V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(a_i-x),
$$
and then set $x=a_1$ to get
$$
V(a_1,a_2,\ldots,a_n)
=V_{n-1}(a_2,\ldots,a_n)\prod_{i=2}^n(a_i-a_1),
$$
Part (e) applies induction on $n$ to $V_{n-1}(a_2,\ldots,a_n)$ (the starting case is $V_0()=1=\prod_{1\leq i<j\leq 0}1$, an empty product, or if you fear $n=0$ it is $V_1(a)=1=\prod_{1\leq i<j\leq 1}1$, still an empty product), to get
$$
V(a_1,a_2,\ldots,a_n)
=\left(\prod_{2\leq i<j\leq n}(a_j-a_i)\right)\prod_{j=2}^n(a_j-a_1)
=\prod_{1\leq i<j\leq n}(a_j-a_i).
$$
Hint:
Consider $$D =
\begin{vmatrix}
1&x&x^2&\cdots&x^{n-2} & x^{n-1}&x^n\\
1&a_1&a_1^2&\cdots&a_1^{n-2} & a_1^{n-1}&a_1^n\\
1&a_2&a_2^2&\cdots&a_2^{n-2} & a_2^{n-1}&a_2^n\\
\vdots\\
1&a_n&a_n^2&\cdots&a_n^{n-2} & a_n^{n-1}&a_n^n\\
\end{vmatrix}$$
This is a Vandermonde determinant, so you already know how to calculate it. Look for the coefficient of $x^{n-1}$. On the other hand develop the determinant using the first row.
In a similar way, you can see the following generalization:
$$\begin{vmatrix}
1&a_1&\cdots&a_1^{k-1}&a_1^{k+1}\cdots &a_1^n\\
1&a_2&\cdots&a_2^{k-1}&a_2^{k+1}\cdots &a_2^n\\
\vdots\\
1&a_n&\cdots&a_n^{k-1}&a_n^{k+1}\cdots &a_n^n\\
\end{vmatrix} = \sigma_{n-k}(a_1,a_2\cdots,a_n)\prod_{i<j}(a_j-a_i)$$
(Here $\sigma_k$ denotes the $k$-th elementary symmetric polynomial)
Best Answer
Subtract the last row from each preceding row to obtain
$$D_n(a_1, \ldots, a_n;x) = \left|\begin{matrix} -x&a_2&\cdots&a_{n}\\ a_1&-x&\cdots&a_{n}\\ \vdots&\vdots&\ddots&\vdots\\ a_1&a_2&\cdots&-x \end{matrix}\right| = \left|\begin{matrix} -x-a_1&0&\cdots&0&a_{n}+x\\ 0&-x-a_2&\cdots&0&a_{n}+x\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&-x-a_{n-1}&a_n+x\\ a_1&a_2&\cdots&a_{n-1}&-x \end{matrix}\right|$$
Now expand along the first column and iterate the procedure: $$= (-x-a_1)\left|\begin{matrix} -x-a_2&\cdots&0&a_{n}+x\\ \vdots&\ddots&\vdots&\vdots\\ 0&\cdots&-x-a_{n-1}&a_n+x\\ a_2&\cdots&a_{n-1}&-x \end{matrix}\right|+(-1)^{n+1}a_1\left|\begin{matrix} 0&\cdots&0&a_{n}+x\\ -x-a_2&\cdots&0&a_{n}+x\\ \vdots&\ddots&\vdots&\vdots\\ 0&\cdots&-x-a_{n-1}&a_n+x\\ a_2&\cdots&a_{n-1}&-x \end{matrix}\right|$$ \begin{align} &=(-x-a_1)D_{n-1}(a_2, \ldots, a_n;x)+a_1(-x-a_2)\cdots(-x-a_{n-1})\\ &=(-x-a_1)(-x-a_2)D_{n-2}(a_3, \ldots, a_n;x)+a_1(-x-a_2)\cdots(-x-a_{n-1})+a_2(-x-a_1)(-x-a_3)\cdots (-x-a_n)\\ &=\cdots\\ &=(-x-a_1)\cdots(-x-a_n)+\sum_{i=1}^n a_i(-x-a_1)\cdots(-x-a_{i-1})(-x-a_{i+1})\cdots (-x-a_n) \end{align}
which is the same as @user1551's result.