So here is the thing. I post here my answer to 312004, since that one is closed; while also insisting this one here has its answer in one oscillator: as usual in QFT, the infinity of oscillators is only a smokescreen to test the student's understanding of the second quantization notation, just part b) of your text's problem. In an ideal world, this FT question above should be merged with the closed one, parts a) and c).
And no, field eigenstates are not an integral part of the QFT toolbox, given the output expected of QFT, at least in particle physics, the focus of your text. Nevertheless, Schroedinger wave functional theory (Jackiw, 1988 and Lüscher, 1985) still floats around.
(Coherent states are more commonly used, e.g. Itzykson and Zuber's QFT text, eqn (3-65) et seq.)
Essentially answering 312004 :
This is the obverse/completion of the twin question 292899. It hinges on a point not uncommon in squeezed-states and string discussions.
I’ll just race to the answer suitably tweaking standard techniques, e.g., of Fischer, Nieto, & Sandberg, 1984, and leaving you to patch up numerical normalizations to your satisfaction.
Recall $[a , a^\dagger ]=1$.
$$
|x\rangle\equiv n(x) ~ \exp\left ( -(a^\dagger -x\sqrt{2})^2 /2 \right )|0\rangle \Longrightarrow \hat{~x}~|x\rangle=x|x\rangle,\\
|p\rangle\equiv n(p) ~ \exp((a^\dagger +ip\sqrt{2})^2 /2 ) |0\rangle \Longrightarrow \hat{~p}~|p\rangle=p|p\rangle ~.
$$
You may try to fix $n(x)=e^{x^2/2}/\pi^{1/4}$ from $\langle 0|x\rangle$, the Schrödinger ground state of the oscillator. (Yes, indeed, n is the inverse of the Gaussian!)
Your assignment is to normal-order
$$
\langle x|p\rangle= n(x) n(p) \langle 0| e^{-(a-x\sqrt{2})^2/2}
e^{(a^\dagger +ip\sqrt{2})^2/2} |0\rangle~.
$$
That is to say, commute all $a$s to the right, where they annihilate the vacuum, and so disappear themselves, leaving the x’s, etc, and the same for the creation operators on the vacuum to the left, depositing their ps.
Actually, given this task, it is easier to normal order, instead, the new oscillators
$$
b\equiv a-x\sqrt{2}, \qquad b^\dagger\equiv a^\dagger +ip\sqrt{2}, \qquad [b,b^\dagger]=1.
$$
Thus,
$$
b|0\rangle= -x\sqrt{2} |0\rangle, \qquad \langle0|b^\dagger=\langle 0| ip\sqrt{2} ,
$$
so that the normal-ordered expression will net <0|0> times a function of x cancelling n(x), same for p, and the $e^{ixp}$ sought.
The answer (derived by standard moves in the Appendix below) is
$$
e^{-b^2/2} e^{b^{\dagger 2}/2}= e^{b^{\dagger 2}/4 }e^{-\ln 2 (1/2+b^\dagger b) }e^{-b^2/4}=\\
e^{b^{\dagger 2}/4 } \left ( \sum^\infty_{k=0}
\frac{(-1/2)^k}{k!} b^{\dagger k} b^k \right )e^{-b^2/4} /\sqrt{2} .
$$
Consequently, the v.e.v. of the above operator is
$$
\langle p |x\rangle= e^{-p^2/2} e^{ixp}e^{-x^2/2} n(x) n(p) /\sqrt{2\pi}= e^{ixp}/ \sqrt{2\pi}.
$$
To prove orthogonality, e.g. of the position states, well, insert a complete set of momentum eigenstates in $\langle x|y\rangle$, and do the dp integral of the two plane waves you just found, netting you a $\delta(x-y)$.
Appendix. The crucial lemmata in the derivation are:
Defining
$$
L_-\equiv -b^2/2, \qquad L_+ \equiv b^{\dagger 2 }/2, \qquad L_0\equiv \tfrac{1}{2} ( 1/2 +b^\dagger b),
$$
note they close into the algebra of SU(1,1) (underlying the Virasoro algebra),
$$
[L_+, L_-]=2 L_0, \qquad [L_0, L_{\pm} ]=\pm L_{\pm}.
$$
The key point: A group element identity (product of exponentials of generators) holds for all representations, but, conversely, an identity of such group elements in a faithful representation, such as the doublet (the Pauli matrices), cannot hold if the generic abstract one does not. (This is nontrivial: it requires Poincaré's exponential theorem, to the effect that the CBH expression found in the exponent is fully in the Lie algebra, and hence the representation is immaterial.) In this simplest rep, then,
$$
L_+= \begin{pmatrix}
0&1\\
0&0
\end{pmatrix} , \qquad
L_-= \begin{pmatrix}
0&0\\
1&0
\end{pmatrix} , \qquad
2L_0= \begin{pmatrix}
1&0\\
0&-1
\end{pmatrix} .
$$
It is then evident that
$e^{L_-} e^{L_+} =e^{L_+/2} e^{-\ln 2 \cdot 2L_0} e^{L_-/2} $ ,
since
$$
\begin{pmatrix}
1&0\\
1&1
\end{pmatrix} \begin{pmatrix}
1&1\\
0&1
\end{pmatrix} = \begin{pmatrix}
1&1/2\\
0&1
\end{pmatrix} \begin{pmatrix}
1/2&0\\
0&2
\end{pmatrix} \begin{pmatrix}
1&0\\
1/2&1
\end{pmatrix} .
$$
However, the middle group element, $ e^{-\ln 2 ~(1/2+b^\dagger b)} $, is not normal ordered yet, but straightforward to get there, given the identity
$$
\bbox[yellow,5px]{e^{c b^\dagger b}= \sum^\infty_{k=0} \frac{(e^c-1)^k}{k!} b^{\dagger k} b^k }.
$$
(Just write out the first four powers of c in the Wick theorem ascending expansion in normal-ordered bs. This is attributed to Wilcox (1967) JMP 8 962-982, and ultimately McCoy, 1932, but it is all but manifest. The oscillator propagator.)
The shortest way to see it is by exploiting the algebraic isomorphism $b^\dagger =x, ~~b=\partial$; so that the left-hand side acting on f(x) is just scaling, $\exp(c x\partial) ~f(x)= f(e^c x)$; while the right-hand one is the Taylor series around x, shifted by $x(e^c-1)$, and hence $f(x)\mapsto f(x+x(e^c-1))=f(e^c x)$.
Back to field theory: all you need to do now is consider an infinity of oscillators, whence, loosely, $\hat{x}\mapsto \hat{\phi}(x),\quad \hat{p}\mapsto \hat{\pi}(x)$, construct the eigenstates as in 292899, and generalize the above to $\langle \Pi|\Phi\rangle $, mutatis mutandis.
As a straightforward consequence, observe the translationally invariant state in the kernel of the momentum is $$
\left|\Pi=0\right\rangle =e^{\int d^{3}x ~ \hat{\phi}_{+} (\vec{x})^{2}/2}\left|0 \right\rangle .
$$
