The zeros of the function are those values of [tex]x[/tex] that make [tex]y=0[/tex]. So we solve
[tex]0=\dfrac{1+50\sin x}{x^2+3}[/tex]
The denominator will always be positive, so we can multiply both sides of the equation by it to get
[tex]0=1+50\sin x\implies \sin x=-\dfrac1{50}[/tex]
[tex]\implies x=\arcsin\left(-\dfrac1{50}\right)+2n\pi=-\arcsin\dfrac1{50}+2n\pi[/tex]
where [tex]n[/tex] is any integer. If we take [tex]n=\pm1[/tex] we should get the two solutions immediately adjacent to the one near [tex]x=0[/tex] that still lie in the interval [tex]-5\le x\le5[/tex]. So the other two zeros are [tex]x=-\arcsin\dfrac1{50}\pm\pi[/tex].
The tangent line to the curve at any [tex]x[/tex] is determined by the value of the derivative of the function at that value of [tex]x[/tex]. So first compute the derivative:
[tex]y=\dfrac{1+50\sin x}{x^2+3}\implies y'=\dfrac{50\cos x(x^2+3)-2x(1+50\sin x)}{(x^2+3)^2}=\dfrac{(50x^2+150)\cos x-100x\sin x-2x}{(x^2+3)^2}[/tex]
Now just plug in the values of [tex]x[/tex] determined above. It's helpful to note
[tex]\cos\left(\arcsin\dfrac1{50}\right)=\dfrac{7\sqrt{51}}{50}[/tex]
[tex]\cos(\theta+2n\pi)=\cos\theta[/tex]
[tex]\sin(\theta+2n\pi)=\sin\theta[/tex]
