You need to find the height which is AO from the triangle ADO having one side = 6 and one angle =42 degrees. Using the tangent of the angle and cross multiply, you will get:
Tan ADO = AO/OD
Height of the pyramid: AO = OD Tan ADO = 6 Tan 42 = 5.4
Now you need the area of the base which is a square and is two times the area of the isosceles right triangle BDC.
Area of the triangle BDC = Base x height / 2 = 12 x 6 / 2 = 36
Area of the square FBCD is twice of it which is 2 x 36 = 72
Volume of AFBCD = Area of the base x height / 3 = 72 x 5.4 / 3 = 129.6
To get the Surface Area you need to find the side of the base and the Slant Height of the pyramid. Perimeter of the base x Slant Height / 2.
The side of the base is calculated from the right triangle ODC which the hypothenuse is 6 x 2 = 12. So to find its leg X:
X^2 + X^2 = 12^2
2X^2 = 144
X^2 = 144/2 and X = 12 Radical 2 / 2 = 6 Radical 2 which is CD.
To find the Slant Height we need to draw the height of the triangle ADC which is AK and join O to K. So
OK = 6 radical 2 /2 = 3 radical 2
AK^2 = AO^2 + OK^2 = 5.4^2 + (3 Radical 2)^2
AK=6.8673
Surface Area = Perimeter of the base x Slant height / 2 = 4 x 6 Radical 2 x 6.8673 / 2 = 116.54
