1. Express \( z \) from the second equation and substitute into the first:

\[
\begin{aligned}
& z=a\left(x+2 y+\frac{5}{2}\right) \rightarrow x^{2}+y^{2}+2 x-y+a\left(x+2 y+\frac{5}{2}\right)=0 \rightarrow \\
& \rightarrow\left(x+\frac{a+2}{2}\right)^{2}+\left(y+\frac{2 a-1}{2}\right)^{2}=\frac{(a+2)^{2}}{4}+\frac{(2 a-1)^{2}}{4}-\frac{5 a}{2}
\end{aligned}
\]

There is a unique solution if \(\frac{(a+2)^{2}}{4}+\frac{(2 a-1)^{2}}{4}-\frac{5 a}{2}=0 \rightarrow(a-1)^{2}=0 \rightarrow a=1\).

For the parameter \(a=1\), the solution is \(x=-\frac{a+2}{2}=-\frac{3}{2}, y=-\frac{2 a-1}{2}=-\frac{1}{2}, z=x+2 y+\frac{5}{2}=0\).
Answer: 1) $a=1$; 2) $x=-\frac{3}{2}, y=-\frac{1}{2} ; z=0$