There are various adaptations of the simplex method, which can be tedious and labour-intensive. And it's easy to make mistakes! The graphical method I find to be easier to understand and less prone to error. However, if there are more than two variables, the graphical method becomes difficult or even possible and the simplex method is the better method.
Let's look at the graphical method for 35 and 36 first. Graphing inequalities consists of drawing graphs of the given lines as if they were equalities. The inequalities are represented by shading areas to one side or the other of the lines, depending on what the inequality is. In linear programming we are usually only interested in "first quadrant" graphing, that is, where the variables are both positive (x1,x2≥0).
35) -2x1+x2≤4 is plotted as -2x+y=4, which has x intercept at x=-2 and y intercept at y=4. Draw a line through these points (remember we are only interested in the first quadrant. The line y=10 is horizontal. (For graphing purposes x1 is replaced by x and x2 by y. It doesn't matter which is which, the final result of the linear programming will be the same.)
For -2x1+x2≤4 the area to be shaded is on the right of the line -2x+y=4. To decide where to shade, simply substitute positive values for x and y and if they satisfy the inequality, then the point (x,y) is in the shaded region. For example, if x=y=1 then we have -2+1=-1 which is less than 4 so (1,1) is in the shaded region. For x2≤10 the shaded region is below the line y=10. The feasible region is where the shadings overlap. However, the feasible region is an open area and has no upper limits for maximising the objective function (profit) p=2x1+3x2. For example, let x1=100 and x2=10 then the inequalities are satisfied and p=230. x1 can be increased to any value and will always satisfy the inequalities. That means the question has been wrongly stated (min p?). Given the information the simplex method won't work.
If the question had been to minimise p (in this sort of question it would have been min c), then there is a solution: we need to evaluate the two vertices of the feasible region (3,10), giving p=6+30=36 and (0,4), giving p=0+12=12, so min p is when x1=0 and x2=4. There is no point using the simplex method because there may be an error in the question.
36) Unfortunately, we have the same error in this question as we had in 35). There is no limit because the feasible region is unbounded. Please check your question and also check the constraints in the question you submitted a few days ago--there is something missing.