Solving this is a matter of simple arithmetic and some logical thinking. 
This is another Tough Nut that you should be able to crack with the following ideas to help you.
<p/> 
Each of the numbers 1 to 15 is used to label one
of the vertices or one of the edges of this tree.
<p/>
For the vertex magic investigation, a graph is said to be vertex magic if the sum of
the numbers on a vertex and on all the edges
joined to that vertex is the same for each vertex 
and we call this the magic sum. Let's suppose
the magic sum is 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>h</mi></mrow></math> and this is the same at
each vertex. The total of all the numbers 1 + 2 +
... + 15 = 120 but the numbers on the edges, say

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>a</mi><mo>,</mo><mi>b</mi><mo>,</mo><mi>c</mi><mo>,</mo><mi>d</mi><mo>,</mo><mi>e</mi><mo>,</mo><mi>f</mi><mo>,</mo><mi>g</mi></mrow></math> are each counted twice, and as there are 8 vertices, so
<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
    <mstyle displaystyle="true"><mrow><mo stretchy="false">(</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>+</mo><mi>c</mi><mo>+</mo><mi>d</mi><mo>+</mo><mi>e</mi><mo>+</mo><mi>f</mi><mo>+</mo><mi>g</mi><mo stretchy="false">)</mo><mo>+</mo><mn>120</mn><mo>=</mo><mn>8</mn><mi>h</mi><mo>.</mo></mrow>
    </mstyle></math>
</td></tr></table>
<br />


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So 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>a</mi><mo>+</mo><mi>b</mi><mo>+</mo><mi>c</mi><mo>+</mo><mi>d</mi><mo>+</mo><mi>e</mi><mo>+</mo><mi>f</mi><mo>+</mo><mi>g</mi></mrow></math> is a multiple of 8 and it is
at least 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mn>1</mn><mo>+</mo><mn>2</mn><mo>+</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>+</mo><mn>7</mn><mo>=</mo><mn>28</mn></mrow></math> so it must be 32 or more.
Hence 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mn>8</mn><mi>h</mi><mo>&#x2267;</mo><mn>152</mn></mrow></math> and 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>h</mi><mo>&#x2267;</mo><mn>19</mn></mrow></math>. So you can try 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>h</mi><mo>=</mo><mn>19</mn></mrow></math> or 20 or 21 or ...
<p/>
For the edge magic investigation, a graph is said to be edge magic if, for each edge,
the sum of the three numbers on the edge and the two
vertices at its ends is the same and this is called 
the magic constant. Let's call this 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>k</mi></mrow></math>. How do we discover what edge magic graphs there
are? The three order-3 vertices 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>a</mi><mo>,</mo><mi>b</mi><mo>,</mo><mi>c</mi></mrow></math> are
counted three times (once for each edge at that vertex) and the sum of all the
numbers is again 120.
<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
    <mstyle displaystyle="true"><mrow><mn>2</mn><mo stretchy="false">(</mo><mi>a</mi><mo>+</mo><mi>b</mi><mo>+</mo><mi>c</mi><mo stretchy="false">)</mo><mo>+</mo><mn>120</mn><mo>=</mo><mn>7</mn><mi>k</mi><mo>.</mo></mrow>
    </mstyle></math>
</td></tr></table>
<br />

Hence 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>k</mi></mrow></math> is even and since 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>a</mi><mo>+</mo><mi>b</mi><mo>+</mo><mi>c</mi><mo>&#x2267;</mo><mn>6</mn></mrow></math> we have

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mn>7</mn><mi>k</mi><mo>&#x2267;</mo><mn>120</mn><mo>+</mo><mn>12</mn><mo>=</mo><mn>132</mn></mrow></math> and so 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>k</mi><mo>&#x2267;</mo><mn>20</mn></mrow></math>.

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