You are a bit confused. The slope of the tangent is not merely $\frac{f(x)-f(a)}{x-a},$ it's the limit as $x\to a$ of that fraction.
In this case, the point in question is $(1,3)$, which means that $x=1$, and that $f(1)=3$. So you are trying to find $\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{(4x-x^2)-3}{x-1} = \lim_{x\to 1}\frac{-(x^2-4x+3)}{x-1}.$ Now, unsurprisingly, the numerator and denominator both evaluate to $0$ at $x=1$. Why "unsurprisingly"? Because this always happens when you try to compute the slope of the tangent. That's why we use limits. This being a rational function, you can always factor out $x-a$ from the numerator (in this case, $x-1$); so it was not that you "got lucky and accidentally found" that you could factor, with rational functions like this, a polynomial divided by a polynomial, that's what you should always be looking for. Indeed, $-(x^2-4x+3) = -(x-3)(x-1),$ so $\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{-(x^2-4x+3)}{x-1} = \lim_{x\to 1}\frac{-(x-1)(x-3)}{x-1} = \lim_{x\to 1}-(x-3),$ and this limit can be evaluated simply by plugging in $x=1$.
Definition 2 is even easier. We want to find $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},$ where $a=1$. Now, $f(1) = 3$, we knew that already. What is $f(1+h)$? Plug in! $f(1+h) = 4(1+h) - (1+h)^2 = 4+4h - (1+2h+h^2) = 4+4h-1-2h-h^2 = 3+2h-h^2.$ So we have: $\lim_{h\to 0}\frac{f(1+h)-f(1)}{h} = \lim_{h\to 0}\frac{(3+2h-h^2)-3}{h} = \lim_{h\to 0}\frac{2h-h^2}{h}.$ Why is this easier? Because here it should be obvious how to factor the numerator so you can cancel it with the $h$ in the denominator and simplify: you have $2h-h^2 = h(2-h)$. Cancel, and then do the resulting easy limit.