Let f(x) = (3x²+x−5)/(x²+4). Find f(3), f(2), f(1), f(a+1), f(sqrt(a)), f(a²), f(a)+f(1).

Step 1: Solve question 1.
Given $f(x) = \frac{3x^2 + x - 5}{x^2 + 4}$.

a) Find $f(3)$.

$$f(3) = \frac{3(3)^2 + 3 - 5}{(3)^2 + 4} = \frac{3(9) + 3 - 5}{9 + 4} = \frac{27 + 3 - 5}{13} = \frac{25}{13}$$

The value is $\boxed{\frac{25}{13}}$.

b) Find $f(2)$.

$$f(2) = \frac{3(2)^2 + 2 - 5}{(2)^2 + 4} = \frac{3(4) + 2 - 5}{4 + 4} = \frac{12 + 2 - 5}{8} = \frac{9}{8}$$

The value is $\boxed{\frac{9}{8}}$.

c) Find $f(1)$.

$$f(1) = \frac{3(1)^2 + 1 - 5}{(1)^2 + 4} = \frac{3(1) + 1 - 5}{1 + 4} = \frac{3 + 1 - 5}{5} = \frac{-1}{5}$$

The value is $\boxed{-\frac{1}{5}}$.

d) Find $f(a + 1)$.

$$f(a+1) = \frac{3(a+1)^2 + (a+1) - 5}{(a+1)^2 + 4}$$ $$f(a+1) = \frac{3(a^2 + 2a + 1) + a + 1 - 5}{a^2 + 2a + 1 + 4}$$ $$f(a+1) = \frac{3a^2 + 6a + 3 + a - 4}{a^2 + 2a + 5}$$ $$f(a+1) = \frac{3a^2 + 7a - 1}{a^2 + 2a + 5}$$

The value is $\boxed{\frac{3a^2 + 7a - 1}{a^2 + 2a + 5}}$.

e) Find $f(\sqrt{a})$.

$$f(\sqrt{a}) = \frac{3(\sqrt{a})^2 + \sqrt{a} - 5}{(\sqrt{a})^2 + 4}$$ $$f(\sqrt{a}) = \frac{3a + \sqrt{a} - 5}{a + 4}$$

The value is $\boxed{\frac{3a + \sqrt{a} - 5}{a + 4}}$.

f) Find $f(a^2)$.

$$f(a^2) = \frac{3(a^2)^2 + a^2 - 5}{(a^2)^2 + 4}$$ $$f(a^2) = \frac{3a^4 + a^2 - 5}{a^4 + 4}$$

The value is $\boxed{\frac{3a^4 + a^2 - 5}{a^4 + 4}}$.

g) Find $f(a) + f(1)$.

$$f(a) + f(1) = \frac{3a^2 + a - 5}{a^2 + 4} + \left(-\frac{1}{5}\right)$$ $$f(a) + f(1) = \frac{5(3a^2 + a - 5) - 1(a^2 + 4)}{5(a^2 + 4)}$$ $$f(a) + f(1) = \frac{15a^2 + 5a - 25 - a^2 - 4}{5a^2 + 20}$$ $$f(a) + f(1) = \frac{14a^2 + 5a - 29}{5a^2 + 20}$$

The value is $\boxed{\frac{14a^2 + 5a - 29}{5a^2 + 20}}$.

Step 2: Solve question 2.
Given $g(x) = \frac{x}{3x+2}$ and $h(x) = \sqrt{x^2 - 1}$.

a) Find $g(1) + h(2)$.

$$g(1) = \frac{1}{3(1)+2} = \frac{1}{5}$$ $$h(2) = \sqrt{(2)^2 - 1} = \sqrt{4 - 1} = \sqrt{3}$$ $$g(1) + h(2) = \frac{1}{5} + \sqrt{3}$$

The value is $\boxed{\frac{1}{5} + \sqrt{3}}$.

b) Find $g(3) h(1)$.

$$g(3) = \frac{3}{3(3)+2} = \frac{3}{9+2} = \frac{3}{11}$$ $$h(1) = \sqrt{(1)^2 - 1} = \sqrt{1 - 1} = \sqrt{0} = 0$$ $$g(3) h(1) = \frac{3}{11} \times 0 = 0$$

The value is $\boxed{0}$.

c) Find $\frac{g(5)}{h(2)}$.

$$g(5) = \frac{5}{3(5)+2} = \frac{5}{15+2} = \frac{5}{17}$$

From part (a), $h(2) = \sqrt{3}$.

$$\frac{g(5)}{h(2)} = \frac{\frac{5}{17}}{\sqrt{3}} = \frac{5}{17\sqrt{3}}$$

Rationalize the denominator:

$$\frac{5}{17\sqrt{3}} = \frac{5}{17\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{17 \times 3} = \frac{5\sqrt{3}}{51}$$

The value is $\boxed{\frac{5\sqrt{3}}{51}}$.

d) Find $g(a - 1) + h(a + 1)$.

$$g(a-1) = \frac{a-1}{3(a-1)+2} = \frac{a-1}{3a-3+2} = \frac{a-1}{3a-1}$$ $$h(a+1) = \sqrt{(a+1)^2 - 1} = \sqrt{a^2 + 2a + 1 - 1} = \sqrt{a^2 + 2a}$$ $$g(a-1) + h(a+1) = \frac{a-1}{3a-1} + \sqrt{a^2 + 2a}$$

The value is $\boxed{\frac{a-1}{3a-1} + \sqrt{a^2 + 2a}}$.

e) Find $g(a^2) h(a^2)$.

$$g(a^2) = \frac{a^2}{3a^2+2}$$ $$h(a^2) = \sqrt{(a^2)^2 - 1} = \sqrt{a^4 - 1}$$ $$$$