**Basic approach.** Use the same hypergeometric distribution you used for the first question to determine
$$ p_k \stackrel{\text{def}}{=} P(\text{out of the $5$ chosen printers, exactly $k$ are bad}) $$
Note then that also $p_k = P(\text{repairs cost $50k$ dollars})$. So you can easily determine expected total cost $C$ as
$$ E(C) = \sum_{k=0}^4 50k p_k $$
Find $E(C^2)$ in a similar way, and then
$$ \sigma^2_C = E(C^2)-[E(C)]^2 $$