Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.
Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).
Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.