Yes. Consider a two player game described by the following matrix

\begin{array}{|c|c|c|} \hline & L & R \\ \hline a & 3,0 & 0,0 \\ \hline b & 0,0 & 3,0 \\ \hline c & 2,0 & 2,0 \\ \hline \end{array}

If only pure strategies are allowed, only $a$ and $b$ are rationalizable, they are best responses to $L$ and $R$ respectively, while $c$ is not best response to anything. But if you look at the mixed extension of the game $c$ strictly dominated 0.5$a$+0.5$b$, hence 0.5$a$+0.5$b$ is not rationalizable. And as a result in the mixed extension all $a$, $b$ and $c$ are rationalizable.

If you insist on an example with three players I can add a third one to this. This third player has exactly one possible strategy, so it does not really matter what his payoffs are and what he does.
If you think that is cheating, the third player can have multiple strategies that do not at all influence the payoff of the first two players.

Yes. Consider a two player game described by the following matrix

\begin{array}{|c|c|c|} \hline & L & R \\ \hline a & 3,0 & 0,0 \\ \hline b & 0,0 & 3,0 \\ \hline c & 2,0 & 2,0 \\ \hline \end{array}

If only pure strategies are allowed, only $a$ and $b$ are rationalizable because they are best responses to $L$ and $R$ respectively while $c$ is not a best response to anything. But if you look at the mixed extension of the game then $c$ strictly dominates 0.5$a$+0.5$b$, hence 0.5$a$+0.5$b$ is not rationalizable. As a result, in the mixed extension all $a$, $b$ and $c$ are rationalizable.

If you insist on an example with three players I can add a third one to this. This third player has exactly one possible strategy, so it does not really matter what his payoffs are and what he does.
If you think that is cheating, the third player can have multiple strategies that do not at all influence the payoff of the first two players.

There two question. The first one is

If $a$ and $b$ are two pure rationalizable strategies, can 0.5$a$+0.5$b$ fail to be a rationalizable strategy?

Yes. Consider the example given

\begin{tabular}{c|c|c|c|} \multicolumn{1}{c}{} & \multicolumn{1}{c}{$b_1$} & \multicolumn{1}{c}{$b_2$} & \multicolumn{1}{c}{$b_3$} \\ \cline{2-4} $a_1$ & 6,9 & 6,3 & 7,2 \\ \cline{2-4} $a_2$ & 5,4 & 5,6 & 5,5 \\ \cline{2-4} $a_3$ & 8,3 & 0,2 & 2,2 \\ \cline{2-4} \end{tabular}

$$ A(h) \cdot a(0) = \left( \begin{array}{ccc} 1 \\ 1 \\ 1+h \end{array} \right), $$

Yes. Consider a two player game described by the following matrix

\begin{array}{|c|c|c|} \hline & L & R \\ \hline a & 3,0 & 0,0 \\ \hline b & 0,0 & 3,0 \\ \hline c & 2,0 & 2,0 \\ \hline \end{array}

If only pure strategies are allowed, only $a$ and $b$ are rationalizable, they are best responses to $L$ and $R$ respectively, while $c$ is not best response to anything. But if you look at the mixed extension of the game $c$ strictly dominated 0.5$a$+0.5$b$, hence 0.5$a$+0.5$b$ is not rationalizable. And as a result in the mixed extension all $a$, $b$ and $c$ are rationalizable.

If you insist on an example with three players I can add a third one to this. This third player has exactly one possible strategy, so it does not really matter what his payoffs are and what he does.
If you think that is cheating, the third player can have multiple strategies that do not at all influence the payoff of the first two players.