How do you go about solving these?

<p><a href="http://i.imgur.com/APNKU.jpg%5B/url%5D"&gt;http://i.imgur.com/APNKU.jpg&lt;/a&gt;&lt;/p>

<p><a href="http://i.imgur.com/SpIDT.jpg%5B/url%5D"&gt;http://i.imgur.com/SpIDT.jpg&lt;/a&gt;&lt;/p>

<ol>
<li>If x=0, then a^x is 1. Thus, ka^x=k=1/2
Thus, the function is y=(1/2)a^x
Plugging in x=1,
2=(1/2)a^1
2=a/2
a=4</li>
</ol>

<p>Draw a line from h to a vertex on the square base. You form a right triangle with legs e and m/sqrt(2)(Since the diagonal of a square is sqrt2 times its side, the distance is m(sqrt2)/2, or m/sqrt2). e^2=(m/sqrt2)^2+h^2(Pythagorean Tm). e=m, so
h^2=m^2-(m^2)/2=0.5m^2
Thus, h=m/sqrt2</p>