Recurrence relation (linear, second-order, constant coefficients) Q1. Find the general solution to the difference equation
$$ a_{n} - 4a_{n-1} + 3a_{n-2} = 6 $$  
Q2. Solve the difference equation 
$$ a_{n} - a_{n-1} - 2a_{n-2} = 0, a_0 = 2, a_1 = 4 $$  
I am completely lost in solving recurrence relation questions. Can anyone guide me in steps to solve the following 2 questions? 

Simple, general way of disposing such equations is using generating functions.
Define:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
Write the recurrence so it hasn't subtractions in indices:
$$
a_{n + 2} - 4 a_{n + 1} + 3 a_n = 6
$$
Multiply by $z^n$, sum over $n \ge 0$ an recognize the resulting terms:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} - 4 \frac{A(z) - a_0}{z} + 3 A(z) = 6 \frac{1}{1 - z}
$$
Substituting your initial values, and solving for $A(z)$, writing that as partial fractions:
$$
A(z) = \frac{3}{(1 - z)^3} + \frac{5}{2 (1 - z)} + \frac{5}{2 (1 - 3 z)}
$$
Using the generalized binomial theorem you can read off the coefficients:
\begin{align}
a_n &= 3 \binom{-3}{n} (-1)^n + \frac{5}{2} + \frac{5}{2} \cdot 3^n \\
    &= 3 \binom{n + 3 - 1}{3 - 1} + \frac{5}{2} + \frac{5}{2} \cdot 3^n \\
    &= \frac{3 (n + 2) (n + 1)}{2} + \frac{5}{2} + \frac{5}{2} \cdot 3^n \\
    &= \frac{5 \cdot 3^n + 3 n^2 + 9 n + 11}{2}
\end{align}
The other one I leave as a exercise for the gentle reader.
