a) From the Pythagorean theorem, we have
AC² = AB² + BC² = a² + a² ⇒ AC = √2 a
and
AK² = AB² + BK² = a² + (2a)² ⇒ AK = √5 a
ABCD and BKLC are perpendicular, so AC and CL are perpendicular and ∆ACL is a right triangle. By the Pythagorean theorem, we have
AL² = AC² + CL² = (√2 a)² + (2a)² ⇒ AL = √6 a
Now,
AK² + LK² = (√5 a)² + a² = 6a² = (√6 a)² = AL²
which means ∆AKL satisfies the Pythagorean theorem and is a right triangle, so ∠AKL = 90°.
b) We know AC = √2 a and AK = √5 a. By the Pythagorean theorem, we find
CK² = BC² + BK² = a² + (2a)² ⇒ CK = √5 a
By the law of cosines, we have
AK² = AC² + CK² - 2 AC CK cos(∠AKC)
6a² = 2a² + 5a² - 2√10 a² cos(∠AKC)
cos(∠AKC) = 1/(2√10)
∠AKC = arccos(1/(2√10)) ≈ 80.9°