For the first question, I think the answer is no. Consider the following example:
$X = Spec k[x,y,z]/(xy - z^2)$ a quadric cone. Consider the Cartier divisor $D = V(z)$. It has two irreducible components corresponding to the ideals $(x,z)$ and $(y,z)$ respectively (these are non-Cartier, Q-Cartier divisors). Both components smooth and they meet at the origin and so the multiplicity of $D$ (of this nodal singularity) is $2$. By multiplicity, I assume you mean the multiplicity of the scheme $D$ at a point.
On the other hand, if you blow up the origin $(x,y,z)$ you get a chart $$k[x/z, y/z, z]/( (x/z)(y/z) - 1).$$ The pull back of $D$ on this chart is just $z = 0$ (one copy of the exceptional divisor) so the order of $\mu^*(D)$ along $E$ is equal to 1. (The order of the components along the exceptional divisor is $1/2$ in each case, but they are $\mathbb{Q}$-Cartier)
There's a deeper problem in your first question though. If I recall correctly, in general, when you blow-up a point $x \in X$ on a singular variety, there isn't a unique prime exceptional divisor lying over $x$. There are probably multiple such divisors. To make matters worse, the pull back of your given Cartier divisor can have different multiplicities along these different exceptional divisors.
For the second question:
You assume that the pair $(X, \Delta)$ is klt, and you define the discrepancy at $E$ to be the order along $E$ of $K_Y - \mu^*(K_X + \Delta)$. Then you say that you know that $a(E, X, \Delta) \leq 1$ if $X$ is smooth. This isn't true.
I assume you know that the definition of klt implies that these discrepancies are all $> -1$. However, consider the following example.
$X = Spec k[x,y,z]$ and $\Delta = 0$. This pair is certainly klt. When you blow up the origin though, the relative canonical divisor $K_{Y/X} = 2E$, two copies of the exceptional (if you blow up the origin in $\mathbb{A}^n$, you get $n-1$ copies of the exceptional divisor). If you blow up points on that exceptional divisor (and repeat), you get further exceptional divisors with greater and greater discrepancy.
Hopefully I didn't misunderstand the question.
Best Answer
Let me start with a little nitpicking:
With that, you can do this:
Let $H$ be a very ample Cartier divisor on $X$ (Homework: why is there such a divisor?). Then $\mu_*H$ is a Weil divisor on $Y$ and if $Y$ is $\mathbb Q$-factorial, then it is $\mathbb Q$-Cartier and hence $\mu^*\mu_*H$ makes sense and is a $\mathbb Q$-Cartier divisor.
Then it is easy to see that $H-\mu^*\mu_*H$ is a $\mu$-very ample $\mathbb Q$-Cartier divisor and also that it is negative effective $\mu$-exceptional. Finally, since $L$ is ample it follows that $a\mu^*L+H-\mu^*\mu_*H$ is ample for $a\gg 0$. (One could for example argue that $aL-\mu_*H$ is ample (actually nef is enough) for $a\gg0$ and hence $a\mu^*L+H-\mu^*\mu_*H$ is also ample).