1. Set up the triangle notation

Let $a=BC=7,\quad b=CA=8,\quad c=AB=\alpha.$ Here angle $A$ is between sides $b$ and $c$.

Given: $\cos A=\frac{2}{3}.$

2. Use the Law of Cosines to find $c$

By the cosine rule, $a^2=b^2+c^2-2bc\cos A.$ Substitute the values: $49=64+c^2-2\cdot 8\cdot c\cdot \frac{2}{3}.$ So, $49=64+c^2-\frac{32c}{3}.$ Multiply by $3$: $147=192+3c^2-32c.$ Hence, $3c^2-32c+45=0.$

Solve: $3c^2-32c+45=(3c-5)(c-9)=0.$ Thus, $c=\frac{5}{3}\quad \text{or} \quad c=9.$ Since $c=\alpha\in \mathbb N$, we get $c=9.$

3. Find $\cos C$

Now the sides are $a=7,\quad b=8,\quad c=9.$ Using the cosine rule at angle $C$, $\cos C=\frac{a^2+b^2-c^2}{2ab}.$ So, \cos C=\frac{49+64-81}{2\cdot 7\cdot 8}=rac{32}{112}=rac{2}{7}.

4. Find $\cos 3C$ using the triple-angle formula

We use $\cos 3C=4\cos^3 C-3\cos C.$ Since $\cos C=\frac{2}{7}$, $\cos 3C=4\left(\frac{2}{7}\right)^3-3\left(\frac{2}{7}\right).$ Compute: 4\left(\frac{8}{343}\right)-\frac{6}{7}=rac{32}{343}-\frac{294}{343}=-\frac{262}{343}.

5. Evaluate $49\cos(3C)+42$

$49\cos 3C+42=49\left(-\frac{262}{343}\right)+42.$ Since $\frac{49}{343}=\frac{1}{7}$, $49\cos 3C+42=-\frac{262}{7}+42.$ Now, $42=\frac{294}{7},$ so $49\cos 3C+42=\frac{294-262}{7}=\frac{32}{7}.$

Thus, $\frac{m}{n}=\frac{32}{7}$ with $\gcd(32,7)=1$. Therefore, $m+n=32+7=39.$

6. Comparison with stored answer

Derived answer: $39$

Stored correct answer: $39$

They match.