Problem 1. (a) Read the definition of isometry in Brown. Your answer should be a complete sentence starting something like this -- "An isometry is a transformation that ... " and not "An isometry is when ..." for an isometry is a thing (an abstract thing) not a "when". (b) An isometry maps a triangle ABC to a congruent triangle A'B'C; by SSS, so angle ABC is congruent to angle A'B'C', etc.
Problem 2. The problem is to construct MN(P). This figure was drawn deliberately so that N(P) is off the page. The solution is that MN is a translation in a direction and distance that can be determined from m and n. One can then translate P without computing N(P) as an intermediate step. Moral. MN is a single isometry T; there are lots of ways of writing it as a product or just as translation determined by a vector. You have a lot of choice and control of how you construct T(P).
Problem 3. This problem can be solved by mirror adjustment. If you are having trouble seeing this while you are preparing for the test, it is very instructive to actually construct T(P) for several P and see what the pattern is. In particular, try constructing T(T(P)). This should be P itself.
The mirror adjustment solution involves writing R as a double line reflection UV, where the second line reflection U is chosen to cancel with m, i.e., the line for the second line reflection U should be m itself. Then T = V.
Problem 4. U is the product of 6*3 = 18 line reflections. Since 18 is an even number, U can be the identity, a translation, a rotation, or a point reflection (which is a rotation).
Problem 5. F is a translation with translation vector (2, -6). So the lines should be perpendicular to the line h through (0,0) and (2,-6). You can pick the first line n to go through any point; the simplest choice would be a line through (0,0). The second line would then go through the midpoint Q of (0,0) and (2,-6).
So this would do the job.