Here's a plan of the view on the ground:

$\displaystyle \setlength{\unitlength}{3.5mm}

\begin{picture}(30,18)

\put(2,0){\line(0,1){15}}

\put(2,0){\line(5,3){19.5}}

\put(2,15){\line(5,-3){12.6}}

\put(2,15){\line(6,-1){19.5}}

\put(2,15){\line(3,-5){6.6}}

\put(8,4.8){\line(5,3){1.1}}

\put(9.5,4.5){\line(-3,5){0.62}}

\put(0.5,0){$O$}

\put(0.5,14.5){$T'$}

\multiput(6,1.4)(6,3.6){3}{$a$}

\put(2.5,1){$\theta$}

\put(0.5,7){$x$}

\put(14.5,6){$P$}

\put(21.5,10.5){$Q$}

\end{picture}$

Let $\displaystyle OT' = x$ (*T'* denotes the base of the tower, so I'm using x to denote the horizontal distance from *O* to the base of the tower, *not* the inclined distance to the top of the tower). Then $\displaystyle h/x = \tan\alpha$. From the picture, $\displaystyle a/x = \cos\theta$. Therefore $\displaystyle \frac ah = \frac{a/x}{h/x} = \frac{\cos\theta}{\tan\alpha}$, and so $\displaystyle h = a\tan\alpha\sec\theta$.

Next, $\displaystyle \frac h{QT'} = \tan\beta$, so $\displaystyle QT'\,^2 = h^2\cot^2\beta = a^2\tan^2\alpha\cot^2\beta\sec^2\theta$. But by Pythagoras (see the above diagram), $\displaystyle QT'\,^2 = a^2(4+\tan^2\theta).$ Thus $\displaystyle \tan^2\alpha\cot^2\beta\sec^2\theta = 4+\tan^2\theta$, and I guess you can take it from there to get the formula for tan^2(θ). Once you have that, the remaining two parts of the question are easy.