Let $\alpha \in {\Bbb C}$, and let $\vert{n}\rangle $ be the harmonic oscillator state with energy $(n+\textstyle\frac{1}{2})\hbar\omega$. At $t=0$, the coherent state $\vert {\alpha(0)}\rangle $ is defined by
$$
\vert{\alpha(0)}\rangle= e^{-\vert \alpha \vert^2/2}\,\left(
\sum_{n=0}^{\infty} \displaystyle{\alpha^n\over \sqrt{n!}}\,\vert{n}\rangle\right) \tag{1}
$$
What is $\vert{\alpha(t)}\rangle$, the coherent state at time $t$? Start with (1). Since $\left\vert n\right\rangle $ is an eigenstate of the harmonic oscillator hamiltonian $\hat{H}=\left( \hat a^{\dagger }\hat a+\frac{1}{2}\right)
\hbar \omega $ with eigenvalue $\left( n+\frac{1}{2}\right) \hbar \omega ,$ the time evolution of $\left\vert n\right\rangle $ is simply $\left\vert
n(t)\right\rangle =e^{-i(n+\frac{1}{2})\omega t}\left\vert n\right\rangle $ and thus
\begin{equation}
\left\vert \alpha (t)\right\rangle =e^{-\left\vert \alpha \right\vert
^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{%
1}{2})\omega t}\left\vert n\right\rangle \right) .
\end{equation}
It is easy to show that $\left\vert \alpha (t)\right\rangle $ is normalized.
Now we first need to show that $a\vert{\alpha(t)}\rangle=\alpha e^{i\hbar \omega
t}\vert{\alpha(t)}\rangle$. Recall that $\hat{a}\left\vert n\right\rangle
=\sqrt{n}\left\vert
n-1\right\rangle .$ \ Then, since $\hat{a}$ is linear,
\begin{eqnarray}
\hat{a}\left\vert \alpha (t)\right\rangle &=&e^{-\left\vert \alpha
\right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}%
e^{-i(n+\frac{1}{2})\omega t}\hat{a}\left\vert n\right\rangle \right) ,
\\
&=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{%
\alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{1}{2})\omega t}\sqrt{n}\left\vert
n-1\right\rangle \right) , \\
&=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{%
\alpha ^{n}}{\sqrt{\left( n-1\right) !}}e^{-i(n+\frac{1}{2})\omega
t}\left\vert n-1\right\rangle \right) , \\
&=&\alpha e^{-i\omega t}e^{-\left\vert \alpha \right\vert ^{2}/2}\left(
\sum_{n=0}^{\infty }\frac{\alpha ^{n-1}}{\sqrt{\left( n-1\right) !}}%
e^{-i(n-1+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) .\quad
\end{eqnarray}
The sum properly starts at $n=1$ since the $n=0$ term does not exist. Thus, setting $m=n-1,$ we can rewrite this sum in terms of $m,$ with $m$ starting at $m=0.$ Hence
\begin{eqnarray}
\hat{a}\left\vert \alpha (t)\right\rangle &=&\alpha e^{-i\omega t}\left[
e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{m=0}^{\infty }\frac{%
\alpha ^{m}}{\sqrt{m!}}e^{-i(m+\frac{1}{2})\omega t}\left\vert
m\right\rangle \right) \right] \\
&=&\alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle .
\end{eqnarray}
A useful secondary result, which follows immediately from above, is
\begin{eqnarray}
\left[ \hat{a}\left\vert \alpha (t)\right\rangle \right] ^{\dagger }
&=&\left\langle \alpha (t)\right\vert \hat{a}^{\dagger } \\
&=&\left[ \alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle \right]
^{\dagger }=\alpha ^{\ast }e^{i\omega t}\left\langle \alpha (t)\right\vert
\end{eqnarray}
Now $\langle \hat p(t) \rangle$ and $\langle \hat x(t)\rangle$ for $\vert{\alpha(t)}\rangle$. Starting from the definitions
$$
\hat{a} =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}+\frac{i}{%
m\omega }\hat{p}\right) , \qquad
\hat{a}^{\dagger } =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}-%
\frac{i}{m\omega }\hat{p}\right) ,
$$
we have
$$
\hat{x} =\sqrt{\frac{\hbar }{2m\omega }}\left( \hat{a}^{\dagger }+%
\hat{a}\right) , \qquad
\hat{p} =i\sqrt{\frac{m\omega \hbar }{2}}\left( \hat{a}^{\dagger }-%
\hat{a}\right) ,
$$
and thus
\begin{eqnarray}
\left\langle x(t)\right\rangle &=&\sqrt{\frac{\hbar }{2m\omega }}\left[
\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha
(t)\right\rangle +\left\langle \alpha (t)\right\vert \hat{a}\left\vert
\alpha (t)\right\rangle \right]\, , \\
&=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha
e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha
(t)\right\rangle \\
&=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha
e^{-i\omega t}\right] ,
\end{eqnarray}
which is real, as expected. We can clean this up by writing $\alpha
=\left\vert \alpha \right\vert e^{i\theta }$ to obtain%
\begin{equation}
\left\langle x(t)\right\rangle =\sqrt{\frac{2\hbar }{m\omega }}\left\vert
\alpha \right\vert \cos \left( \omega t-\theta \right) \tag{2}
\end{equation}
Likewise,
\begin{eqnarray}
\left\langle p(t)\right\rangle &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[
\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha
(t)\right\rangle -\left\langle \alpha (t)\right\vert \hat{a}\left\vert
\alpha (t)\right\rangle \right] \\
&=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \alpha ^{\ast }e^{i\omega
t}-\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert
\alpha (t)\right\rangle \\
&=&-\sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left( \omega
t-\theta \right) \tag{3}
\end{eqnarray}
which is again real.
In your specific case you are starting with a coherent state for which, at $t=0$, we have
$$
\langle x(0)\rangle= b\sqrt{2}x_0\, ,\qquad \langle p(0)\rangle=0
$$
so this implies from (2) and (3) evaluated at $t=0$ that
$$
b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }}\left\vert
\alpha \right\vert \cos \left(\theta \right)\, , \qquad 0= \sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left(\theta \right)
$$
Comparing with your initial conditions gives $\theta=0$ and $b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }}
\alpha $ with $\alpha$ real.
Finally, $\hat{x}^{2}$ and $\hat{p}^{2}.$ From $\hat{x}$ and $\hat{p},$ we find
\begin{eqnarray}
\hat{x}^{2} &=&\frac{\hbar }{2m\omega }\left( \hat{a}^{\dagger }+
\hat{a}\right) ^{2}=\frac{\hbar }{2m\omega }\left( \left( \hat{a}
^{\dagger }\right) ^{2}+\hat{a}^{\dagger }\hat{a}+\hat{a}\hat{a}
^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\
&=&\frac{\hbar }{2m\omega }\left( \left( \hat{a}^{\dagger }\right) ^{2}+2
\hat{a}^{\dagger }\hat{a}+1+\left( \hat{a}\right) ^{2}\right) , \\
\hat{p}^{2} &=&-\frac{m\omega \hbar }{2}\left( \hat{a}-\hat{a}
^{\dagger }\right) ^{2}=-\frac{m\omega \hbar }{2}\left( \left( \hat{a}
^{\dagger }\right) ^{2}-\hat{a}^{\dagger }\hat{a}-\hat{a}\hat{a}%
^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\
&=&-\frac{m\omega \hbar }{2}\left( \left( \hat{a}^{\dagger }\right) ^{2}-2%
\hat{a}^{\dagger }\hat{a}-1+\left( \hat{a}\right) ^{2}\right) ,
\end{eqnarray}
where
\begin{equation}
\hat{a}\hat{a}^{\dagger }=\hat{a}\hat{a}^{\dagger }-\hat{a}%
^{\dagger }\hat{a}+\hat{a}^{\dagger }\hat{a}=\left[ \hat{a},%
\hat{a}^{\dagger }\right] +\hat{a}^{\dagger }\hat{a}=1+\hat{a}%
^{\dagger }\hat{a}
\end{equation}
has been used. Thus,
\begin{eqnarray}
\left\langle x^{2}(t)\right\rangle &=&\frac{\hbar }{2m\omega }\left[
\left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right)
^{2}\left\vert \alpha (t)\right\rangle +2\left\langle \alpha (t)\right\vert
\hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\
&&\left.\qquad\qquad\qquad\qquad\qquad\quad
+1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha
(t)\right\rangle \right] , \\
&=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}\right)
^{2}+2\alpha ^{\ast }\alpha +1+\left( \alpha e^{-i\omega t}\right) ^{2}%
\right]\, ,\\
&=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega
t}+\alpha e^{-i\omega t}\right) ^{2}+1\right] , \\
&=&\frac{\hbar }{2m\omega }\left[ 4\left\vert \alpha \right\vert ^{2}\cos
^{2}\left( \omega t-\theta \right) +1\right] . \\
\left\langle p^{2}(t)\right\rangle &=&-\frac{m\omega \hbar }{2}\left[
\left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right)
^{2}\left\vert \alpha (t)\right\rangle -2\left\langle \alpha (t)\right\vert
\hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\
&&\left.\qquad\qquad\qquad\qquad\qquad\quad
-1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha
(t)\right\rangle \right] , \\
&=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega
t}\right) ^{2}-2\alpha ^{\ast }\alpha -1+\left( \alpha e^{-i\omega t}\right)
^{2}\right]\, ,\\
&=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast
}e^{i\omega t}-\alpha e^{-i\omega t}\right) ^{2}-1\right] , \\
&=&-\frac{m\omega \hbar }{2}\left[ -4\left\vert \alpha \right\vert ^{2}\sin
^{2}\left( \omega t-\theta \right) -1\right]\, ,\\
&=&\frac{m\omega \hbar }{2}\left[
4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) +1%
\right] .
\end{eqnarray}
I am not entirely sure what you mean with the definition of a coherent state since $e^{ika^\dagger}$ is an operator not a state, but the proof works without it. Also, you didn't define what the coefficients $\alpha$ are in the third step of your calculation.
In the second step, where you used the Baker-Campbell-Hausdorff formula you also missed a minus sign, since one negative sign comes from $i^2$, while one comes from the formula itself, such that
\begin{equation}
e^{ik(a+a^\dagger)} = e^{ika} e^{ika^\dagger} e^{k^2[a,a^\dagger]/2}.
\end{equation}
This means your expectation value is actually
\begin{equation}
\langle 0|e^{ikx}|0\rangle = e^{k^2/2}\langle0|e^{ika}e^{ika^\dagger}|0\rangle.
\end{equation}
You may then use the exponential series, i.e. $e^{x} = \sum_n x^n/n!$, so you end up with
\begin{equation}
\langle 0|e^{ikx}|0\rangle = e^{k^2/2}\sum_{n,m}\frac{1}{n!}\frac{1}{m!}(ik)^n(ik)^m\langle0|a^n(a^\dagger)^m|0\rangle.
\end{equation}
From there, things are pretty straightforward. You just need to show that you require $m=n$ so then you can compute the expectation value.
It might also be worth noting that
\begin{equation}
\langle0|x^2|0\rangle = \langle0|(a+a^\dagger)^2|0\rangle = 1.
\end{equation}
Remark: I dropped all factors in $x$ during my calculations, i.e. I used $x=(a + a^\dagger)$. Obviously, using the right definition of $x$ will change the result, but only by some factor.
Best Answer
There are many ways to go around this. You can start from the coherent states and apply the unitary $\hat{O}$ directly on them. That will not be that simple because you will get a term $\hat{O}e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}$. Now, the typical approach would be to exchange the order of the operators to get something like $e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}\hat{O}$ (up to some extra term due to the commutator). This not really a simple task but once you are done, you can Taylor expand the operator $\hat{O}$ and keep only the zeroth order (all other terms contain annihilation operators acting on vacuum). I am not going to dig into the calculation in more detail, there are many ways to do it and none of them is really pleasant.
But there is a better way. You can define a displacement operator by the action $\hat{D}(\alpha)|0\rangle = |\alpha\rangle$ and then you have $\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha) = \hat{a}+\alpha$. You can combine this with the formulas for $\hat{O}\hat{a}\hat{O}^\dagger$, $\hat{O}\hat{b}\hat{O}^\dagger$ to see how the annihilation operators are transformed. What you should get is a beam-splitting of the two coherent states, i.e., $$|\alpha,\beta\rangle\to|t\alpha+r\beta,t\beta-r\alpha\rangle$$, where $t = \cos\theta$, $r = \sin\theta$.