See attached file.
To find the tangent vector you just differentiate the components of the vector w.r.t. the parameter and then normalize the result. Let's denote the position vector by V. Then if:
V = [lambda, (lambda-1)^2, - lambda]
dV/dlambda = [1, 2(lambda-1), - 1]
at point P the parameter lambda = 1, so the derivative there is [1, 0, - 1]. The tangent vector is obtained by normalizing this to 1: 1/sqrt(2) [1, 0, - 1]
V = [cos(mu), sin(mu), mu-1]
dV/d mu = [-sin(mu), cos(mu), 1]
At P the parameter mu = 0, so there the derivative is [0,1,1] and the ...
A detailed solution is given.