You are only allowed to ask one question at a time. I am providing the result for tan(arc cos (2/3)).

Let arc cos (2/3) = y

=> tan y = sin y/ cos y

arc cos (2/3) = y

=> cos y = (2/3)

sin y = sqrt (1 - (cos y)^2)

=> sin y = sqrt (1 - (2/3)^2)

=> sin y = sqrt ((9 - 4)/9)

=> sin y = (sqrt 5)/3

tan y = [(sqrt 5)/3]/(2/3)

=> (sqrt 5)/2

**This gives tan (arc cos (2/3) = (sqrt 5)/2**

Since the arccosine function returns the vaue of the angle, we'll write arccos (2/3) = t => cos t = 2/3

We'll write the tangent function as a fraction:

tan t = sin t/ cos t

tan t = sin (arccos (2/3))/cos (arccos (2/3))

We know that cos (arccos x) = x and sin (arccos x) = sqrt (1 - x^2)

tan t = sqrt(1 - 4/9)/(2/3)

tan t = sqrt5/3/(2/3)

We'll simplify and we'll get:

tan t = (sqrt 5)/2

**The required value of tan (cos ^-1(2/3)) is (sqrt 5)/2.**