(α) We have
$$\begin{aligned} k^4 + 4 &= (k^2)^2 + 4k^2 + 2^2 - 4k^2 = (k^2 + 2)^2 - (2k)^2 \\ &= (k^2 + 2 - 2k)(k^2 + 2 + 2k) = [(k-1)^2 + 1^2][(k+1)^2 + 1^2]. \end{aligned}$$

(β) We multiply both terms of the fraction by $(2^4)^n$, to receive:
$$\begin{aligned} K &= \frac{\left(2^4 + \frac{1}{4}\right)\left(4^4 + \frac{1}{4}\right)\left(6^4 + \frac{1}{4}\right) \cdots \left[(2n)^4 + \frac{1}{4}\right]}{\left(1^4 + \frac{1}{4}\right)\left(3^4 + \frac{1}{4}\right)\left(5^4 + \frac{1}{4}\right) \cdots \left[(2n-1)^4 + \frac{1}{4}\right]} \\ &= \frac{(3^2 + 1)(5^2 + 1)(7^2 + 1)(9^2 + 1)(11^2 + 1) \cdots [(4n-3)^2 + 1][(4n-1)^2 + 1][(4n+1)^2 + 1]}{(1^2 + 1)(3^2 + 1)(5^2 + 1)(7^2 + 1)(9^2 + 1)(11^2 + 1)(13^2 + 1) \cdots [(4n-3)^2 + 1][(4n-1)^2 + 1]} \\ &= \frac{(4n+1)^2 + 1}{1^2 + 1} = 8n^2 + 4n + 1 = 4n^2 + 4n^2 + 4n + 1 = (2n)^2 + (2n+1)^2. \end{aligned}$$