Harmonic Motion Problems: Solving for Maximum Speed and Velocity after Collision

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1. A 66.3 g mass is attached to the end of
an unstressed vertical spring (of constant
63.5 Nm) and then dropped.
The acceleration of gravity is 9.8 m/s2 :
What is its maximum speed? Answer in
units of m/s.

Im not quite sure that I understand what this problem is saying. Is the block hanging from the spring? Is it on top of the spring? I am not sure how this system is setup, but beyond that it is a harmonic motion problem, correct?

2.A(n) 1.6 kg object moving at a speed of
6.5 m/s strikes a(n) 1.2 kg object initially
at rest. Immediately after the collision, the
1.6 kg object has a velocity of 0.88 m/s di-
rected 46 degrees from its initial line of motion.
What is the speed of the 1.2 kg object
immediately after the collision? Answer in
units of m/s.

Ive been doing it like this:
X direction: m1i*v1i=m1i*v1f*cos(46) + m2*v2f*cos(46)

Y direction: m1*v1i=m1*v1f*sin(46) + m2*v2f*sin(46)

and then v2f= sqrt(x^2 + y^2)

However this is apparently wrong, am I close in my approach to this problem?
 
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I believe the other end of the spring is attached to a ceiling. If that is the case, then you can apply Newton's second law to a point where the mass has moved a distance x from it's equilibrium position.

mg - kx = m d^2 x/ dt^2 and solve for dx/dt. That will lead you to the maximum velocity.
 
X direction: m1i*v1i=m1i*v1f*cos(46) + m2*v2f*cos(46)
Y direction: m1*v1i=m1*v1f*sin(46) + m2*v2f*sin(46)

You have assumed that mass m2 travels in the same direction as m1 after the collision. This assumption is wrong. Assume that m2 travell in a direction theta with respect to the original direction of m1.

m1i*v1i=m1i*v1f*cos(46) + m2*v2f*cos(theta)
0 =m1*v1f*sin(46) - m2*v2f*sin(theta)

solve for theta.