You can tackle many tricky geometry questions with a single word ‘Construction’. Yes, just precisely draw the diagram as asked in the question and manually measure the unknown side/angle that has been asked…

Disclaimer: Use this method only when you are unable to figure out how to solve a question.

Q. 1) If BE and CF are two medians of a triangle ABC and G is the centroid and EF and AG intersects each other at O then find – AO:OG
(A) 1:1             (B) 1:2              (C) 2:1                (D) 3:1

Since no information about the type of triangle is given, hence we will assume it to be ‘Equilateral’.
Step 1: Draw an Equilateral triangle of any size(but it should be big enough for the purpose). Take 6 cm.
How to draw an equilateral triangle? Just draw a line AB = 6 cm, and make ∠A = 60 and ∠B = 60, the point where the two angles will intersect, will be point C.
Step 2: Now draw the medians BE and CF. Make sure that E and F lie exactly in the middle of AC and AB respectively.
Step 3: Complete the remaining figure.
Step 4: Measure AO and OG with a ruler.

You will find that AO:OG = 3:1

Q. 2)

Again, a simple question.
Step 1: Draw a circle of any radius with centre O and diameter AB.
Step 2: Put protractor at A and draw ∠CAB = 34.
Step 3: Join BC
Step 4: Measure ∠CBA with protractor.

Q. 3)

This may appear tricky if you go on solving it with conventional methods. But it is way too easy when solved through construction.
Step 1: Draw a triangle ABC with AC = 6 cm. Here we are using the concept of scaling. In the exam you can’t draw a line 12 cm long. Hence we are reducing AC to 1/2. Hence, when we will measure AE, we will have to multiply it by 2 to get the exact answer.
Step 2: Draw internal bisector of ∠ABC. Now, be very cautious, a slight mis-measurement give you a wrong answer. Use protractor for accurate measurements (For simplicity, you can take ∠A = 60 and hence use protractor to make ∠CBD = 30)
Step 3: Again with the help protractor, make ∠BDA = 90
Step 4: Draw DE parallel to BC
Step 5: Measure AE with a scale. You will get AE = 3 cm.
Hence actual length of AE = 2*3 = 6 cm
Method:

AD extended meets BC at F.
∠ADB = ∠FDB (BD is the angle bisector)
Triangles ABD and FBD are congruent. So AD = DF
Triangles ADE and AFC are similar