Assume all variables are single letters.

(1)

λx.xz λy.xy
(λx.((xz) (λy.(xy))))

(λx.xz) λy.w λw.wyzx 
((λx.(xz)) (λy.(w (λw.(((wy)z)x)))))

λx.xy λx.yx
(λx.((xy) (λx.(yx))))

(2)

  (λz.z) (λy.y y)(λx.x a)
= (λy.y y)(λx.x a)
= (λy.y y)(λx.x a)
= ((λx.x a) (λx.x a))
= ((λx.x a) a)
= (a a)

  (λz.z)(λz.z z)(λz.z y)
= (λz.z z)(λz.z y)
= (λz.z y) (λz.z y)
= ((λz.z y) y)
= (y y)

  (λx.λy.x y y)(λa.a) b
= (λy.(λa.a) y y) b
= (λa.a) b b
= (b b)

  ((λf.λa.(f a) λx.x) λs.(s s))
= (λa.(λs.(s s) a) λx.x)

and not

  ((λf.λa.(f a) λx.x) λs.(s s))
= (λa.(λx.x a) λs.(s s))
= (λa.a λs.(s s))
= λs.(s s)

  ((λf.λa.(f a) λs.(s s)) λx.x)
= (λa.(λs.(s s) a) λx.x)
= (λa.(a a) λx.x)
= (λx.x λx.x)
= λx.x


(3)

zero = λs.λz.z
one = λs.λz.(s z)
two = λs.λz.(s (s z))
three = λs.λz.(s (s (s z)))
four = λs.λz.(s (s (s (s z))))
five = λs.λz.(s (s (s (s (s z)))))
...
...
and the definition of addition:
plus = λm.λn.λs.λz.(m s (n s z))
Show that
(plus two three)
reduces to
five

  (plus two three)
= (λm.λn.λs.λz.(m s (n s z)) λs.λz.(s (s z)) λs.λz.(s (s (s z))) )
= 2 beta reductions
  λs.λz.(λs.λz.(s (s z)) s (λs.λz.(s (s (s z))) s z))
= 2 beta reductions
  λs.λz.(s (s (λs.λz.(s (s (s z))) s z)))
= 2 beta reductions
  λs.λz.(s (s (s (s (s z)))))
= five