Confusion in derivation of bias variance decomposition

Question

This question is regarding derivation of bias-variance decomposition as answered (which was accepted) in another thread. I am repeating the steps for this question:

\begin{align} \newcommand{\var}{{\rm var}} {\rm variance} &= \var \left( \frac{1}{k} \sum_i^k Y(x_i) \right) \\ &= \frac{1}{k^2} \sum_i^k \var \left( f(x_i) + \epsilon_i \right) \\ &= \frac{1}{k^2} \sum_i^k \var \left( f(x_i) \right) + \var \left( \epsilon_i \right) \\ &= \frac{1}{k^2} \sum_i^k \var \left( \epsilon_i \right) \\ &= \frac{1}{k^2} k \sigma_\epsilon^2 \\ &= \frac{\sigma^2_\epsilon}{k} \end{align}

Question 1: How did the author get $$\frac{1}{k^2}$$ in the 2nd step?

Answer 1

\begin{align} var(aX)&=E \left[ (aX-E\right[ {aX}\left])^2\right]\\ &=E\left[ (aX-aE\right[X\left])^2\right]\\ &=E\left[ (a(X-E\right[X\left]))^2\right]\\ &=E\left[a^2(X-E\right[X\left])^2\right]\\ &=a^2E\left[ (X-E\right[X\left])^2\right]\\ &=a^2var(X) \end{align} This depends on $E(aX)=aE(X)$.