To make the computations easy to understand, we first consider general equations.
Let $a,b,n$ be integers.
Suppose gcd$(a, n) = 1$.
Consider the following equation.
$ax \equiv b$ (mod $n$)
Since gcd$(a, n) = 1$, we can solve $ay \equiv 1$ (mod $n$) by Euclid's algorithm.
Then $x = by$ (mod $n$) is the solution.
Let $a,b,n,m$ be integers.
Suppose gcd$(n, m) = 1$.
Consider the following equatiuons.
$x \equiv a$ (mod $n$)
$x \equiv b$ (mod $m$)
Since gcd$(n, m) = 1$, we can find, by Euclid's algorithm, integers $k, l$ such that
$mk \equiv 1$ (mod $n$)
$nl \equiv 1$ (mod $m$)
Then $x = amk + bnl$ (mod $nm$) is the solution.
Now let's solve the given equations
$5k \equiv 3$ (mod $6$)
$5k \equiv 2$ (mod $7$)
We get(by Euclid's algorithm or just by testing)
$k \equiv 3$ (mod $6$)
$k \equiv 6$ (mod $7$)
Then we can apply the above method to find $k$.
Since
$7 \equiv 1$ (mod $6$)
$-6 \equiv 1$ (mod $7$)
$k = 3\cdot7 -6\cdot6 = -15 \equiv 27$ (mod $42$)
Here is a way of writing things: $q_i$ and $r_i$ are the quotient and the remainder of the $i$-th division, $u_i$ and $v_i$ are Bézout's coefficients of $r_i$ relative to $76408$ and $345$ respectively.
The recurrence relations are $$u_{i+1}=u_{i-1}-q_iu_i, \quad v_{i+1}=v_{i-1}-q_iv_i.$$
$$ \begin{array}[t]{r@{\qquad}r@{\qquad}r@{\qquad}r}
r_i & u_i & v_i & q_i\\
\hline
76408 & 1 & 0 & \\
345 & 0 & 1 & 221\\
\hline
163 & 1 & -221 & 2\\
19 & -2 & 443 & 8 \\
11 & 17 & -3765 & 1 \\
8 & -19 & 4208 & 1\\
3 & 36 & -7973 & 2\\
2 & -91 & 20154 & 1\\
1 & 127 & -28127 \\
\hline
\end{array}$$
So $\,1=127\cdot 76408-28127\cdot 345$ from which we conclude $$ 345^{-1}\equiv -28127\equiv 48281\mod 76408. $$
Best Answer
Start off by looking at $17^n$ mod 10. (In your case, n will end up being $17^{17}$, but that's way too big to calculate yet.)
$17^0$ ends in a $1$, $17^1$ ends in a $7$, $17^2$ is congruent to $7 \times 7$ so it ends in a $9$, $17^3$ likewise is congruent to $9 \times 7$ so it ends in a $3$, and finally $17^4$ is congruent to $3 \times 7$ so it ends in a $1$.
Since $17^0$ and $17^4$ are congruent mod 10, it follows that $17^n$ mod 10 will repeat every time the exponent $n$ goes up by 4.
Therefore, to solve your problem, you now need to calculate the exponent $17^{17}$ mod 4. Then you can use that along with the pattern I just described to get the final answer. Since this is homework, I'll let you calculate $17^{17}$ mod 4 yourself... hint, use the same idea that I used above!