Given that B=45 degrees, a =2, c=4?

Law of Cosines:
You have SideAngleSide

Find b first using:
b² = a² + c² - 2(axc)cosB°

b² = (2)² + (4)² - (2(2x4)cos45°
b² = 4 + 16 - 2(8)cos45°
b² = 20 - 16cos45°
b² ≈ 8.69
b = √(8.69)
b = 2.96

Then..

(b / sinB) = (c / sinC)

(2.96 / sin45°) = (4 / sinC)

(sin45°)(4) = (sinC)(2.96)

((sin45°)(4) / (2.96) = sinC

Take the inverse...
C° = sin^(-1) (times the entire:)((sin45°)(4) / (2.96))

C° = 72.8°

This is the shape we have:
http://66.49.136.137/zbbo/1.jpg

Now we can simply calculate CH:
CH² + BH² = BC²
CH = BH
2 CH² = BC² = 4
CH = √2/2
BH = √2/2
AH = AB - BH = 4 - (√2/2)

AC² = CH² + AH² = 1/2 + (4 - (√2/2))² = .5 + 10.89 = 11.39
AC = 3.37

C = BCH + ACH
ACH = 45°
BCH = arcsin (HA/AC) = 76°
C = 76° + 45° = 121°

Hope this helps...

Using the formula b^2=a^2+c^2-2abcosB
We can get b:
b^2=2^2+4^2-2(2)(4)cos45
b^2=4+16-16cos45
b=2.95
After knowing b, find C by
c^2=a^2+b^2-2abcosC
4^2=2^2+2.95^2-2(2)(2.95)cosC
you do the rest.

67.5? I'm not sure. I didn't get enough info here.