$f(t,x)$ irreducible, unsolvable over $\mathbb Q(t)$ such that its specialization $f(t_0,t)$ over $\mathbb Q$ is inseparable and (un)solvable. The following theorem can be found in e.g. Lang or Van Der Waerden.

Let $f(t,x)\in\mathbb Q[t,x]$ be irreducible with Galois group $G$ over $\mathbb Q(t)$. If $t\to t_0\in\mathbb Q$ is a specialization such that $f(t_0,x)$ is separable, then the Galois group of $f(t_0,x)$ over $\mathbb Q$ is isomorphic to a subgroup of $G$.

When the assumption on separability of the specialized polynomial $f(t_0,x)$ is dropped, the theorem no longer holds; an explicit counterexample is provided in this answer.
The counterexample constructs a solvable polynomial such that its specialization is solvable, so my question is:

Can we construct an irreducible $f(t,x)$, unsolvable over $\mathbb Q(t)$, such that its rational specialization $f(t_0,x)$ is solvable?
Similarly, can we construct such $f(t,x)$, solvable over $\mathbb Q(t)$, such that its rational specialization is unsolvable over $\mathbb Q$?

 A: Take $f(t,x)\in \Bbb{Q}[t][x]$ monic with roots $b_1,\ldots,b_n\in \overline{\Bbb{Q}[t]}$, let $R=\Bbb{Q}[t,b_1,\ldots,b_n]\subset  \overline{\Bbb{Q}[t]}$, for some $t_0\in \Bbb{Q}$ let $P$ be a maximal  ideal of $R$ containing $t-t_0$,
then $f(t_0,x)\in \Bbb{Q}[x]$ splits completely in $R/(t-t_0)$ thus in $R/P$ and $R/P$ is generated by some roots of $f(t_0,x)$ thus $R/P$ is the splitting field of $f(t_0,x)$.

Then $Gal((R/P)/\Bbb{Q}) = D_P / I_P$
where $D_P$ is the subgroup of $G=Gal(R/\Bbb{Q}[t])$ sending $P$ to itself (thus acting on the quotient too) and $I_P$ the subgroup of $D_P$ acting trivially on $R/P$.


*

*Proof : take a primitive element $\beta\in R$ such that $Frac(R)=\Bbb{Q}(t)(\beta)$, it is the splitting field of $f(t,x)$, let $\alpha \in R/P$ be the reduction of $\beta$, $\Bbb{Q}(\alpha)$ is the splitting field of $f(t_0,x)$. The elements of the Galois groups $G=Gal(\Bbb{Q}(t)(\beta)/\Bbb{Q}(t))$ and $H =Gal(\Bbb{Q}(\alpha)/\Bbb{Q})$ are uniquely defined by where they send $\beta,\alpha$. The reduction $\bmod P$ of $\prod_{g\in G} (Y-g(\beta))\in \Bbb{Q}[t][Y]$ is in $\Bbb{Q}[t][Y]/(t-t_0)=\Bbb{Q}[Y]$ and has $\alpha$ in its roots, thus $\prod_{\sigma \in H} (Y-\sigma(\alpha))$ divides it, which means that for each $\sigma\in H$ there is some $g\in G$ such that $g(\beta)=\sigma(\alpha)\bmod P$. This $g$ permutes the roots of the irreducible factors of $f(t_0,x)$ (since they are polynomials in $\alpha$) thus $g\in D_P$ which proves the claim that $D_P\to H$ is surjective.

If $f(t,x)$ is irreducible then $I_P=\{1\}$ iff $f(t_0,x)$ is separable.
The $G$:solvable $D_P/I_P$:unsolvable case cannot happen. The $G$:unsolvable $D_P/I_P$:solvable case is as easy as taking an unsolvable monic quintic $g(x)\in \Bbb{Q}[x]$ and letting $$f(t,x) = x^5 + t (g(x)-x^5), \qquad f(0,x) = x^5, f(1,x) = g(x)$$
