The equation #f(t)=(t^2;tcos(t-(5pi)/4))# gives you the object's coordinates with respect to time:
#x(t)=t^2#
#y(t)=tcos(t-(5pi)/4)#
To find #v(t)# you need to find #v_x(t)# and #v_y(t)#
#v_x(t)=(dx(t))/dt=(dt^2)/dt=2t#
#v_y(t)=(d(tcos(t-(5pi)/4)))/dt=cos(t-(5pi)/4)-tsin(t-(5pi)/4)#
Now you need to replace #t# with #pi/3#
#v_x(pi/3)=(2pi)/3#
#v_y(pi/3)=cos(pi/3-(5pi)/4)-pi/3 cdot sin(pi/3-(5pi)/4)#
#=cos((4pi-15pi)/12)-pi/3 cdot sin((4pi-15pi)/12)#
#=cos((-11pi)/12)-pi/3 cdot sin((-11pi)/12)#
#=cos(pi/12)+pi/3 cdot sin(pi/12)#
Knowing that #v^2=v_x^2+v_y^2# you find:
#v(pi/3)=sqrt(((2pi)/3)^2+(cos(pi/12)+pi/3 cdot sin(pi/12))^2)#
#=sqrt((4pi^2)/9+cos^2(pi/12)+pi^2/9 cdot sin^2(pi/12)+(2pi)/3 cdot cos(pi/12)sin(pi/12))#
#=1/3sqrt(4pi^2+9cos^2(pi/12)+pisin^2(pi/12)+6picos(pi/12)sin(pi/12))#
