Ifigured out the 1st question but i do not understand the second show me step by step! you!

Nice work so far.  I think if there's a problem with the second part, it's probably
because you swapped the 'x' and 'y' components of the velocity in the first part.

If you just draw the line to complete the triangle in the first picture, it'll help you
see that the vertical (y) component of the velocity is the side opposite the 35.2,
so it's the sine of the angle.  So Vy = -7.5 sin(35.2) = -4.32 m/s.

Similarly, the horizontal (x) component of velocity is adjacent to the angle,
so it's the cosine.  Vx = 7.5 cos(35.2) = 6.13 m/s.

You're doing fine work with the sin and cos, you just swapped the two. 
It's no major big deal.

So now let's look at the second part:

The 'x'-component of velocity doesn't change. because gravity can't do
anything horizontally.  So it's still 6.13 m/s .

They say that the object is now moving 58° below horizontal.  Fine.
Look at the cosine of that 58° ...  it's 0.53 = (6.13 m/s) / (magnitude of velocity) .

Magnitude of velocity = 6.13 / 0.53 = 11.57 m/s .

(Where did all that extra velocity come from ?  The y-component is growing
steadily on account of the acceleration of gravity.)

I just looked at your work on the first part again.  You even have the sin and cos
in the right place !  But there was no need to go over to the other acute angle. 
Just stay with the 35.2 angle and everything will be beautiful.
If for some reason you feel you absolutely must work with the 54.8°, then
the 'x' is opposite it ==> sine, and the 'y' is adjacent to it ==> cosine.
But you're really working with this stuff.  I can tell you know what you're doing.
  

It indicates no change in temperature and the process of matter changing its state