Suppose you want to find a linear combination of $X_1$ and $X_2$ such that
$$
\text{corr}(\alpha X_1 + \beta X_2, X_1) = \rho
$$
Notice that if you multiply both $\alpha$ and $\beta$ by the same (non-zero) constant, the correlation will not change. Thus, we're going to add a condition to preserve variance: $\text{var}(\alpha X_1 + \beta X_2) = \text{var}(X_1)$
This is equivalent to
$$
\begin{align*}
\rho
&= \frac{\text{cov}(\alpha X_1 + \beta X_2, X_1)}{\sqrt{\text{var}(\alpha X_1 + \beta X_2) \text{var}(X_1)}} \\
&= \frac{\alpha \overbrace{\text{cov}(X_1, X_1)}^{=\text{var}(X_1)} + \overbrace{\beta \text{cov}(X_2, X_1)}^{=0}}{\sqrt{\text{var}(\alpha X_1 + \beta X_2) \text{var}(X_1)}} \\&= \alpha \sqrt{\frac{\text{var}(X_1)}{\alpha^2 \text{var}(X_1) + \beta^2 \text{var}(X_2)}}
\end{align*}
$$
Assuming both random variables have the same variance (this is a crucial assumption!) ($\text{var}(X_1) = \text{var}(X_2)$), we get
$$
\rho \sqrt{\alpha^2 + \beta^2} = \alpha
$$
There are many solutions to this equation, so it's time to recall variance-preserving condition:
$$
\text{var}(X_1)
= \text{var}(\alpha X_1 + \beta X_2)
= \alpha^2 \text{var}(X_1) + \beta^2 \text{var}(X_2)
\Rightarrow \alpha^2 + \beta^2 = 1
$$
And this leads us to
$$
\alpha = \rho \\
\beta = \pm \sqrt{1-\rho^2}
$$
UPD. Regarding the second question: yes, this is known as whitening.