Alan

There are an infinite number of solutions to this as you can probably infer from the graph below:

However, if you mean within the range 0 to 360° then use the calculator in the form x = asec(2.5) to find the first one:

$${\mathtt{x}} = {asec}{\left({\mathtt{2.5}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{66.421\: \!821\: \!521\: \!798^{\circ}}}$$

If you look carefully at the graph above you will see that the next one is obtained by subtracting this from 360°

I suspect the last word in the question was meant to be Moon!  Weight is a force.  The mass is the same on Moon and Earth, but the force differs.  

Here we have net force = mass*acceleration.

net force = force of gravity - frictional force (assuming any other forces are negligible).

Assuming the frictional force is constant throughout his slide, we have:

101*9.81 - fricforce = 101*3.2    (where gravitational acceleration = 9.81m/s²)

rearrange to get fricforce:

fricforce = 101*9.81 - 101*3.2 = 101*6.61 N

$${\mathtt{fricforce}} = {\mathtt{101}}{\mathtt{\,\times\,}}{\mathtt{6.61}} \Rightarrow {\mathtt{fricforce}} = {\mathtt{667.61}}$$ Newtons

mass (kg) = force (N) divided by acceleration (m/s²)

I'll let you plug in the numbers here.

Force is mass*acceleration  (force is in Newtons, N, when mass is in kg and acceleration is in m/s²), so 

$${\mathtt{force}} = {\mathtt{896.5}}{\mathtt{\,\times\,}}{\mathtt{2.5}} \Rightarrow {\mathtt{force}} = {\mathtt{2\,241.25}}$$ Newtons 

Weight is a force, given by mass*acceleration of gravity.

On the Earth, gravitational acceleration, g, is about 9.81m/s², so on the Moon it is about 9.81/6 m/s². The weight of a 61.7kg dune buggy on the Moon is therefore 61.7*9.81/6 Newtons

$${\mathtt{weight}} = {\frac{{\mathtt{61.7}}{\mathtt{\,\times\,}}{\mathtt{9.81}}}{{\mathtt{6}}}} \Rightarrow {\mathtt{weight}} = {\mathtt{100.879\: \!5}}$$ Newtons

1Hz is one cycle per second, which, in this case, means one complete wavelength per second. So:

$${\mathtt{wavelength}} = {\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{8}}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{14}}}\right)}} \Rightarrow {\mathtt{wavelength}} = {\mathtt{0.000\: \!000\: \!6}}$$ metres

or 600 nanometres (1 nanometre = 10^-9 metres)

By hand: 3*10⁸/(5*10¹⁴) can be written as 3*10⁸*10^-14/5 = 3*10^8-14/5 = 3*10^-6/5 = 0.6*10^-6 = 600*10^-3*10^-6 = 600*10^-3-6 = 600*10^-9 m, or 600 nanometres. 

These two pictures should help.  Ask for more help if they don't!

 Note: tan is negative in the 2nd and 4th quadrants.

Examples:

$${{log}}_{{\mathtt{2}}}{\left({\mathtt{64}}\right)} = {\mathtt{6}}$$  because 2⁶ = 64

$${{log}}_{{\mathtt{2}}}{\left({\mathtt{10}}\right)} = {\mathtt{3.321\: \!928\: \!094\: \!887\: \!362\: \!6}}$$  because 2^3.321928... = 10

On the scientific calculator on the Home page here, if you hover the cursor over the log key, the lg2 key will appear.  In the Math(Input=Result) window type log(x,2) to get the log to the base 2 of x.

Use the Pythagoras theorem:

$$hypotenuse=\sqrt{adjacent^2+opposite^2}$$