\(a\text{C}_2\text{O}_4^{2-} + b\text{MnO}_4^- + c\text{H}^+ \rightarrow x\text{Mn}^{2+} + y\text{H}_2\text{O} + z\text{CO}_2\)
\(a\) and \(x\) respectively are

The given reaction is a redox reaction in an acidic medium.

The oxidation half-reaction is:
\(\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-\)

The reduction half-reaction is:
\(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)

To balance the number of electrons, multiply the oxidation half-reaction by \(5\) and the reduction half-reaction by \(2\):
\(5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10e^-\)
\(2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}\)

Adding the two balanced half-reactions, we get:
\(5\text{C}_2\text{O}_4^{2-} + 2\text{MnO}_4^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2\)

Comparing this with the given equation \(a\text{C}_2\text{O}_4^{2-} + b\text{MnO}_4^- + c\text{H}^+ \rightarrow x\text{Mn}^{2+} + y\text{H}_2\text{O} + z\text{CO}_2\), we find:
\(a = 5\)
\(b = 2\)
\(c = 16\)
\(x = 2\)
\(y = 8\)
\(z = 10\)

Thus, the values of \(a\) and \(x\) are \(5\) and \(2\) respectively.

Answer: \(5, 2\)