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equals the determinant of the matrix obtained by substituting $a_i$ for $x$, so from the original matrix $a_1$ has been replaced by $a_i$, and as this matrix has its rows $1$ and $i$ identical, its determinant vanishes.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
$$
V(a_1,a_2,\ldots,a_n)
=\left(\prod_{2\leq i<j\leq n}(a_j-a_i)\right)\prod_{j=2}^n(a_j-a_1)
=\prod_{1\leq i<j\leq n}(a_j-a_i).
$$
Call that matrix $A$ and notice that it is a permutation of a circulant matrix,
$$
A = CP
$$
Where $P$ is a permutation matrix with ones on the anti-diagonal, and zeros in all other positions. Then
$$
\det[A] = \det[CP] = \det[C]\det[P]
$$
The determinant of the permutation part can be shown to depend on the size $n$. It can be written as
$$
\det[P] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}
$$
Now $C$ is
$$
\begin{bmatrix}
a_{n} & a_{n-1} & a_{n-2} & \cdots & a_1\\
a_{1} & a_{n} & a_{n-1} & \cdots & a_2\\
a_{2} & a_{1} & a_n & \cdots & a_3\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n-1} & a_{n-2} & a_{n-3} & \cdots & a_{n}\\
\end{bmatrix}
$$
$C$ is a circulant matrix. Define the associated polynomial
$$
f(\omega) = a_n + \sum_{k=1}^{n-1} a_k\omega^k
$$
Then using the product formula on the Wikipedia page for circulant matrices,
$$
\det[C] = \prod_{j=0}^{n-1}f(\omega_j),
$$
where $\omega_j=e^{\frac{2\pi i j}{n}}$ and $i=\sqrt{-1}$. Then the final formula is
$$
\det[A] = (-1)^{\left\lfloor\frac{n}{2}\right\rfloor}\prod_{j=0}^{n-1}\left(a_n + \sum_{k=1}^{n-1} a_k\omega_j^k\right)
$$
Best Answer
Observe that $A_n = E_n - I_n$ where $E_n$ is the matrix with all its entries equal to $1$ and $I_n$ is the identity matrix. So the spectrum of $A_n$ will be the same as the spectrum of $E_n$ translated by $-1$. But the spectrum of $E_n$ is one eigenvalue equal to $n$ and $n-1$ eigenvalues equal to zero. Translate this by $-1$ and you have one eigenvalue equal to $n-1$ and $n-1$ eigenvalues equal to $-1$.