In reference to the second problem I am going to slightly correct ThePerfectHacker... Newton's Second Law states that

where F and a are vectors. So in problem 2 we have three forces acting on a single point. The first step is to set up a coordinate system; when in doubt choose +x to the right and +y upward, which is what I believe the book did. So now we break each force up into components in the x and y directions and then add them. They failed to specify an acceleration so I'm going to assume it is 0 m/s2. (That fits with the book's solution.) The components of a vector of zero length are zero so we can ignore components of a.

In the x direction:

mg is straight down, so has no x-component.

F1 is acting up and to the right at an angle of 30 degrees, so

F1x=F1*cos(30).

F2 is acting up and to the left, so F2x will be negative:

F2x=-F2*cos(45).

Thus

.

In the y direction we have something similar. F1 and F2 are going to have positive y components and mg will be negative. So:

, where mg was given in the first part.

Now we need to solve for F1 and F2. This is a system of 2 equations in 2 unknowns, so we can do it. In Physics the best way to start this is with the substitution method. Solve one of the equations for, say the first equation for F2 in terms of F1 and plug this into the second equation. Then you can solve for F1. etc.

-Dan